If BQP contains NP, does this mean that P=NP?
No, $\mathrm{NP}\subseteq\mathrm{BQP}$ is not known to imply $\mathrm P=\mathrm{NP}$. Even the stronger assumption $\mathrm{NP}\subseteq\mathrm{BPP}$ is not known to yield a deeper collapse than $\mathrm{NP}=\mathrm{RP}$ and $\mathrm{PH}=\mathrm{ZPP^{RP}}=\mathrm{BPP}$; in particular, it is not even known to imply $\mathrm{NP}=\mathrm{coNP}$. (However, all these implications are likely true by virtue of their premises being false.)