Proof on Intersection of Sets

You started the proof with the assumption that $$a\in P(A\cap B)$$ and ended with the conclusion that $$a\in P(A)\cap P(B)$$ You have thus shown that there is an implication $$a\in P(A\cap B)\to a\in P(A)\cap P(B)$$ which just means, since $a$ is arbitrary, that any member of the set $P(A\cap B)$ must also be a member of the set $P(A)\cap P(B)$: this is precisely the definition of a subset. Therefore we can conclude: $$P(A\cap B)\subseteq P(A)\cap P(B)$$