Dear Olimipia, thanks for your answer.
But what exactly do you want? Look at the code below. Mma solved the 1st equation quite quickly for pd, but you get three (already complicated) solutions, and each should be inserted to the remaining equations to be solved further. The result is not really convenient.
Of couse one can solve the 1st equation for pf, giving only two solutions to be inserted in the rest, but is this easier? Not really.
f[x_] := x^a
fl = f[lambda];
dfl = D[f[lambda], lambda];
cd = 12/10;
pn = ((pf*(1 + fl))^(-2) + (pd^(-2)))^(-1/2);
sf = (pf*(1 + fl)/pn)^(-2);
sd = (pd/pn)^(-2);
eqns = {pd == (3*(1 - sd) + 6)/(3*(1 - sd) + 5)*cd,
pf == (3*(1 - sf*(1 + fl)^(-2)) +
2*sf*(1 + fl)^(-2))/(3*(1 - sf*(1 + fl)^(-2)) +
2*sf*(1 + fl)^(-2) - 1)*cf, (1 + fl)^3/dfl ==
10*pn^3*pd^(-3)*(pd - cd)*pf^(-2)}
eq11 = eqns[[1]] // Simplify
solpd = Solve[eq11, pd]
solpf = Solve[eq11, pf]