Is $PP^{(PP^A)} = PP^A$ ? I've looked in the Zoo and it seems it is not true because $PH \subseteq P^{PP}$. Nonetheless I've passed by a paper that appears to have used a positive result. It was in the context that $f : \{0,1\}^* \mapsto \{0,1\}^*$ is a function, computable by probabilistic polynomial time oracle machine $M$ with access to an arbitrary oracle $A$. The authors say and I quote: 


  Its also immediate that $G^f$ ≡
  $G^{M^A}$ is computable by a
  probabilistic polynomial time oracle
  machine with access to $A$.


where $G$ is a probabilistic polynomial time oracle machine.

They do it again also for $S^{N^A,M^A}$, $S,N$ being other probabilistic polynomial time oracle machine.

Update: the paper is "Notions of reducibility between cryptographic primitives" - 2004. It was in the proof of Lemma 1 of the part that "If there exists a fully-black-box reduction from a cryptographic primitive P to cryptographic primitive Q then there exists a relativizing reduction from P to Q as well." The definitions of fully-black-box and relativizing reduction is in the paper. 

The question now, according to Lance Fortnow's answer that they are not equal, does that mean that this is a gap in the proof ?
Richard Beigel gives an oracle $A$ where $P^{NP^A}\not\subseteq PP^A$ which immediately gives $PP^{PP^A}\neq PP^A$.