Does $NP=PP$ collapse the counting hierarchy?
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 $\mathrm{CH=NP}$.