The meaning of relativization
Take $A=NP$, as you requested.  $P^{NP}$ is not necessarily equal to $NP$.

Let me give an example why not.  Consider TAUTOLOGY (given a boolean formula $\varphi$, is it true for all possible assignments to the variables?).  TAUTOLOGY is known to be co-NP-complete.  Therefore, TAUTOLOGY most likely is not in NP, since if TAUTOLOGY were in NP, it would follow that NP = co-NP, which is widely suspected not to be the case.

However, TAUTOLOGY is certainly in $P^{NP}$.  Here's why.  Given an oracle for SAT, we can solve TAUTOLOGY as follows: query whether $\varphi$ is in SAT; then flip the answer (if $\varphi$ is in SAT, answer "no, $\varphi$ is not in TAUTOLOGY"; if $\varphi$ is not in SAT, answer "yes, $\varphi$ is in TAUTOLOGY").  This shows how to solve TAUTOLOGY in polynomial time given an oracle for SAT, and SAT is in NP, so this shows that TAUTOLOGY is in $P^{NP}$.

So TAUTOLOGY is an example of a problem that is in $P^{NP}$, but probably isn't in NP (unless NP = co-NP).  Hopefully this helps illustrate why NP is not the same as $P^{NP}$.

If not, here's a little more intuition.  NP is a particular class.  $P^{NP}$ is the class of problems that can be solved by querying an oracle for NP, possibly many times.   The "many times" part is the key difference between NP and $P^{NP}$.  A (many-one) reduction between two NP problems can only query a NP-oracle once, while $P^{NP}$ allows you to query the oracle as many times as you like.  That seems to be a lot more powerful,  and it's the reason why $P^{NP}$ seems to be a bigger class than NP.