Does $NP=PP$ collapse the counting hierarchy? Suppose $NP=PP$. Then a simple argument shows that $PH^{PP}=NP$. Can we go one step further and get $PP^{PP}=NP$? The simple argument is

Theorem If $NP=PP$ then $PH^{PP}=NP$.

Proof $PP$ is closed under complement (due to Gill), so $NP=coNP=PH$. Take any level of $PH^{PP}$: then $\Sigma_i^{P^{PP}}=\Sigma_i^{P^{NP}}=\Sigma_{i+1}^{P}=NP$. $\square$

One plausible-looking way to get to the desired consequence is by observing that in this world, the interactive proof protocol for the $\textsf{Permanent}$ has been derandomized and de-Merlinized to the point where one message to Arthur has perfect completeness and soundness (as $NP=P^{\#P}$ under the hypothesis). If you can exploit this fact and compute the Permanent in some class that's low for $PP$, such as $UP$ or $BQP$ or $SPP$, we're done. That would give us $NP=PP\implies PP=UP$ (for example), which would immediately give $PP^{PP}=PP^{UP}=PP=NP$.

(This came up in my thesis, where I investigate the hypothesis $QMA=PP$, and it also came up when trying to fix Scott Aaronson's broken theorem $PP\subset BQP_{/qpoly}\implies CH=PP=QMA$, Theorem 5 in Oracles are Subtle but not Malicious).
We have
$$\mathrm{PP^{NP}\subseteq PP^{ModPH}\subseteq P^{PP}},$$
thus by the assumption,
$$\mathrm{PP^{PP}\subseteq PP^{NP}\subseteq P^{PP}\subseteq P^{NP}\subseteq NP}$$
as under the assumption, NP closed under complement. This implies $\mathrm{CH=NP}$.