I wouldn't use InverseFunction in the body of the PDE in this way, since NDSolve is perfectly capable of computing the inverse while solving the PDE. To make this post self-contained, I include a few of the OP details:
\[Sigma]N = 1.78*10^4; d = 0.001; tmax = 3000.0;
inversePureFunction = (1.8996253051848473`/lgti[x,t]*(162.99559471365637` Log[1+8.98898678414097` (1-#1)]-172.98458149779734` Log[#1]))&;
pde1 = D[lgti[x, t], x ] == - \[Sigma]N a[x, t ] lgti[x, t];
initialConditions = {lgti[0,t] == 1, lgti[x, 0] == Exp[- \[Sigma]N x]};
Now, the first step is to convert the InverseFunction into an implicit equation. Here is the implicit equation equivalent:
In[50]:= implicitEquation = Thread[
lgti[x,t] (inversePureFunction[a[x,t]]==t),
Equal
]
Out[50]= 1.89963 (162.996 Log[1+8.98899 (1-a[x,t])]-172.985 Log[a[x,t]])==t lgti[x,t]
The idea is to add a pde for a[x,t] to the NDSolve call. So, we need to come up with a pde for a[x,t] and initial conditions. Using pde1 and implicitEquation, we can solve for D[a[x,t],x]:
In[51]:= asol=Quiet[
Solve[{D[implicitEquation,x],pde1},D[a[x,t],x],D[lgti[x,t],x]],
Solve::bdomv
];
asol//InputForm
Out[52]//InputForm=
{{Derivative[1, 0][a][x, t] -> (938.0600000000017*t*a[x, t]^2*(-1.111247243322715 + 1.*a[x, t])*lgti[x, t])/(-19.244057828963523 + 1.*a[x, t])}}
where I use InputForm to avoid ambiguity (and Quiet the Solve::bdomv message). This results in the following pde for a[x,t]:
In[53]:= pde2=Equal@@asol[[1,1]];
pde2//InputForm
Out[54]//InputForm=
Derivative[1, 0][a][x, t] == (938.0600000000017*t*a[x, t]^2*(-1.111247243322715 + 1.*a[x, t])*lgti[x, t])/(-19.244057828963523 + 1.*a[x, t])
Now, we need initial conditions. When t==0, we have:
In[55]:= implicitEquation/.t->0
Out[55]= 1.89963 (162.996 Log[1+8.98899 (1-a[x,0])]-172.985 Log[a[x,0]])==0
which obviously has the solution:
a[x,0]==1
(although getting WL to prove this isn't easy). When x==0, we can just use the original InverseFunction, since it has no problems when x is 0:
In[56]:= a0t = InverseFunction[inversePureFunction/.x->0/.lgti[0,t]->1][t]
Out[56]= InverseFunction[1.89963 (162.996 Log[1+8.98899 (1-#1)]-172.985 Log[#1])&][t]
Finally, we can use NDSolve:
NDSolve[{pde1,pde2,lgti[x,0]==Exp[-\[Sigma]N x],lgti[0,t]==1,a[x,0]==1,a[0,t]==a0t},{a,lgti},{x,0,d},{t,0,tmax}];
