Language in $PSPACE$ and not necessarily in $P$ if $P=PP$? If $P=PP$ then the counting hierarchy collapses to $CH=P$. Because so many complexity classes are contained in $CH$, this causes most classes to now be contained in $P$. My question is whether this is true for all languages in $PSPACE$, or not.


  What is a candidate language $L\in PSPACE$ such that we do not know that $P=PP$ implies that $L\in P$?
Trivially there will be space-bounded classes $P\ne Space(n)\subsetneq Space(n^2)\subsetneq PSPACE$, and $NSpace(n)\subsetneq NSpace(n^2)\subsetneq PSPACE$, which do not collapse due to the space hierarchy theorem.
More interestingly, the Complexity Zoo lists two other classes that is in PSPACE but not in CH, namely

$QSZK$ (Quantum Statistical Zero Knowledge)
$RG[\#1]$ (single-round referreed games).

Perhaps $QSZK$ satisfies your condition. The best published upper bound (that I can find) is that $QSZK\subseteq PSPACE$ by Watrous[1]. Perhaps it does not satisfy your condition; because it has a complete problem that asks to determine the trace distance between the quantum states produced by two quantum circuits given as input, and this is typically doable in $P^{\#P}$, which you have assumed is equal to $P^{\#P}=P$. I don't know, and invite people to think about whether $QSZK\subseteq CH$ or not. Since you asked specifically for a hierarchy, you can think about the hierarchy $QSZK^{{QSZK}^{\cdots}}$. It is unknown whether this class is self-low, i.e. whether $QSZK=QSZK^{QSZK}$ and the hierarchy collapses.
Update
Known results:
$$QMA=QIP[1]\subseteq A_0PP\subseteq PP$$
$$QMA[2]=QMA[k]$$
$$QSZK\subseteq QIP[2]\subseteq QIP[3]=QIP=PSPACE.$$
Perhaps $RG[\#1]$ satisfies your condition. No upper bound is known except that $RG[\#1]\subseteq RG[\#2]=PSPACE\subseteq RG=EXP$.
[1] J. Watrous. Limits on the power of quantum statistical zero-knowledge, to appear in Proceedings of IEEE FOCS'2002