complexity of proof of p(n) grows greater with n if for all x P(x) is unprovable? Is it true that if "for all x P(x)" is unprovable in pA then the complexity of the proof of P(n) becomes greater as n grows bigger?
 A: You didn’t specify what you mean by “complexity”. If one interprets it as the number of lines in the proof, this is a famous conjecture of Kreisel (usually stated contrapositively: if there is a constant $c$ such that $\phi(\overline n)$ has a PA proof with at most $c$ lines for all $n$, then PA proves $\forall x\,\phi(x)$).
Kreisel’s conjecture is still open for the most natural formulation of PA, however it is extremely sensitive to the choice of the language, and various variants of it have been proved and disproved. See Hrubeš for a recent paper in the positive direction, where you can also find pointers to other known results related to the conjecture.
A: Consider the assertion $P(x)$ that asserts "$x$ is not the Gödel code of a proof of a contradiction in PA".  If PA is consistent, then we can prove $P(n)$ for any particular natural number $n$, since no such $n$ codes a proof. I don't know what is your measure of proof complexity, but the proof of $P(n)$ in every case is mundane: it is because the sequence coded by $n$ violates one of the syntactic requirements of being a proof. Namely, one of the sequents is not an axiom, or does not follow by modus ponens from the earlier sequents, or the conclusion is not a contradiction, and so on. In particular, the formulas appearing in the proofs of $P(n)$ have bounded complexity (although these proofs do get longer, since one must check more cases to cover the earlier sequents). But meanwhile, the assertion $\forall x\, P(x)$ is independent. 
