For the second item, the value for D00, D11, D22, and D33 are fixed. I am only trying to have the combination for P00, P11, P22, and P33 along with the fixed Dxx. So it should have 24 possible combinations of Pxx. For the third item, I am having all the possible combinations for both Pxx and Dxx. The value of Pxx could be any and the value of Dxx could be any. So, there are 32 binary variables to govern them. For both items, Pxx should be greater than Dxx in any combination before running the optimization. But I am taking care of that in the code.

P00 > D00, P11 > D11, P22 > D22, P33 > D33

For your first request, I modify my code and add a Print to show only the last two items.

v=Select[Tuples[{0,1},16],
  Total[Take[#,{1,4}]]==...==1&];
Print[Drop[v,22]];

which displays

{{1,0,0,0,0,1,0,0,0,0,0,1,0,0,1,0},{1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1}}

which corresponds to

{{z0,z5,z11,z14},{z0,z5,z10,z15}}

Is that what you were asking for?

For your second item, do I correctly understand that you are removing the assignments giving values to D00, D11, D22 and D33? If I do that then do I make any changes to the assignments to TC0, TC1, TC2 and TC3 which use D00, D11, D22 and D33? Do I understand that I then use the 8 constraints that follow to create the 24 alternatives as before? If so then I believe the current Select matches your equations you have given for this.

I do not believe I understand your third item? Are there now 32 binary variables instead of 16?

Note: If it might make my code any easier to understand there are two items.

Mathematica starts list indexes at one, not zero. You named your variables starting at zero. So z0 corresponds to #[[1]], z1 corresponds to #[[2]], etc.

Next, you can replace

{P00,P11,P22,P33,D00,D11,D22,D33}=
{z0 P0+z1 P1+z2 P2+z3 P3, z4 P0+z5 P1+z6 P2+z7 P3, z8 P0+z9 P1+z10 P2+z11 P3,
 z12 P0+z13 P1+z14 P2+z15 P3, z0 D0+z1 D1+z2 D2+z3 D3, z4 D0+z5 D1+z6 D2+z7 D3,
 z8 D0+z9 D1+z10 D2+z11 D3, z12 D0+z13 D1+z14 D2+z15 D3}/.r[[i]];

with

P00=z0 P0+z1 P1+z2 P2+z3 P3/.r[[i]];
P11=z4 P0+z5 P1+z6 P2+z7 P3/.r[[i]];
P22=z8 P0+z9 P1+z10 P2+z11 P3/.r[[i]];
P33=z12 P0+z13 P1+z14 P2+z15 P3/.r[[i]];
D00=z0 D0+z1 D1+z2 D2+z3 D3/.r[[i]];
D11=z4 D0+z5 D1+z6 D2+z7 D3/.r[[i]];
D22=z8 D0+z9 D1+z10 D2+z11 D3/.r[[i]];
D33=z12 D0+z13 D1+z14 D2+z15 D3/.r[[i]];

if that is any easier for you to understand.

Can you perhaps adapt this

v=Select[Tuples[{0,1},16],
  Total[Take[#,{1,4}]]==1&&Total[Take[#,{5,8}]]==1&&
  Total[Take[#,{9,12}]]==1&&Total[Take[#,{13,16}]]==1&&
  #[[1]]+#[[5]]+#[[9]]+#[[13]]==1&&#[[2]]+#[[6]]+#[[10]]+#[[14]]==1&&
  #[[3]]+#[[7]]+#[[11]]+#[[15]]==1&&#[[4]]+#[[8]]+#[[12]]+#[[16]]==1&];
r=Map[MapThread[Rule,{{z0,z1,z2,z3,z4,z5,z6,z7,z8,z9,z10,z11,z12,z13,z14,z15},#}]&,v]

which finds the 24 acceptable assignments.

And then this

{P00,P11,P22,P33} = {z0 P0+z1 P1+z2 P2+z3 P3,z4 P0+z5 P1+z6 P2+z7 P3,
  z8 P0+z9 P1+z10 P2+z11 P3,z12 P0+z13 P1+z14 P2+z15 P3}/.r[[1]]
{D00,D11,D22,D33} = {z0 D0+z1 D1+z2 D2+z3 D3,z4 D0+z5 D1+z6 D2+z7 D3,
  z8 D0+z9 D1+z10 D2+z11 D3,z12 D0+z13 D1+z14 D2+z15 D3}/.r[[1]]

uses the first such assignment to find the first set of your eight expressions

When I try to execute your code I get errors, probably because I do not have the definition of ETC or f[x] and possibly other symbols, so I can not experiment to see if small changes might give you better results. I also do not see how your code is making use of the 24 different combinations of P and D.

If you can provide additional information then I will take a few minutes and see what I can find.

I moved your code around a little so that I could try NMinimize on each of the 24 acceptable assignments

K=100; pi=2.5; h=0.4; s=0.1; \[Lambda]=0.2; c1=(h/(h+pi)); c2=1-c1;
P0=1200; P1=950; P2=750; P3=600; D0=850; D1=650; D2=500; D3=400;
\[Alpha]0=0.0022919; \[Alpha]1=0.014506; \[Alpha]2=0.025709; \[Alpha]3=0.045571;
f[x_]:=\[Lambda] Exp[-\[Lambda] x];
v=Select[Tuples[{0,1},16],
  Total[Take[#,{1,4}]]==Total[Take[#,{5,8}]]==Total[Take[#,{9,12}]]==Total[Take[#,{13,16}]]==
    #[[1]]+#[[5]]+#[[9]]+#[[13]]==#[[2]]+#[[6]]+#[[10]]+#[[14]]==
    #[[3]]+#[[7]]+#[[11]]+#[[15]]==#[[4]]+#[[8]]+#[[12]]+#[[16]]==1&];
r=Map[MapThread[Rule,{{z0,z1,z2,z3,z4,z5,z6,z7,z8,z9,z10,z11,z12,z13,z14,z15},#}]&,v];
Table[
{P00,P11,P22,P33,D00,D11,D22,D33}={
  z0 P0+z1 P1+z2 P2+z3 P3, z4 P0+z5 P1+z6 P2+z7 P3, z8 P0+z9 P1+z10 P2+z11 P3,
  z12 P0+z13 P1+z14 P2+z15 P3, z0 D0+z1 D1+z2 D2+z3 D3, z4 D0+z5 D1+z6 D2+z7 D3,
  z8 D0+z9 D1+z10 D2+z11 D3, z12 D0+z13 D1+z14 D2+z15 D3}/.r[[i]];
e0:=Integrate[\[Alpha]0 P00 (t0-x) f[x],{x,0,t0}];
e1:=Integrate[\[Alpha]1 P11 (t1-x) f[x],{x,0,t1}];
e2:=Integrate[\[Alpha]2 P22 (t2-x) f[x],{x,0,t2}];
e3:=Integrate[\[Alpha]3 P33 (t3-x) f[x],{x,0,t3}];
TC0=K D00/(P00 t0)+(h(P00-D00)t0)(pi/(h+pi))/2+s D00 e0/(P00 t0);
TC1=K D11/(P11 t1)+(h(P11-D11)t1)(pi/(h+pi))/2+s D11 e1/(P11 t1);
TC2=K D22/(P22 t2)+(h(P22-D22)t2)(pi/(h+pi))/2+s D22 e2/(P22 t2);
TC3=K D33/(P33 t3)+(h(P33-D33)t3)(pi/(h+pi))/2+s D33 e3/(P33 t3);
ETC=PP0 TC0+PP1 TC1+PP2 TC2+PP3 TC3;
Const={(P11(t1))PP0==(P11(t1)-P11 \[Alpha]1 Integrate[(t1-x)f[x],{x,0,t1}])PP1,
  (P22(t2)+P11(t1)-P11 \[Alpha]1 Integrate[(t1-x)f[x],{x,0,t1}])PP1==
    (P11(t1))PP0+(P22(t2)-P22 \[Alpha]2 Integrate[(t2-x)f[x],{x,0,t2}])PP2,
  (P33(t3)+P22(t2)-P22 \[Alpha]2 Integrate[(t2-x)f[x], {x,0,t2}])PP2==
    (P22(t2))PP1+(P33(t3)-P33 \[Alpha]3 Integrate[(t3-x)f[x],{x,0,t3}])PP3,
  PP0+PP1+PP2+PP3==1, ETC>0, PP0>0, PP1>0, PP2>0, PP3>0, t0>=0.1, t1>=0.1, t2>=0.1, t3>=0.1};
Decision={PP0,PP1,PP2,PP3,t0,t1,t2,t3};
Sol=Flatten[{NMinimize[{ETC,Const},Decision, Method->"NelderMead"],r[[i]]}],
  {i,1,24}]

When I run that it fairly quickly responds with the minimum it found for each of the 24 cases.

None of the minimum are close to the value you are expecting.

There are a few things in the code that worry me a little, but I have not identified a serious problem yet.

Please check this very carefully to see if I might have damaged your code with my changes.

What I'd prefer to see is a complete stand-alone example.

What is being minimized? Is the goal speed, or quality of result?

The values of all z should be either one or zero.