Adleman's theorem to $\mathsf{P=BPP}$
First of all, $\mathrm{NP}\subset\mathrm{P}/log$ would imply that $\mathrm{NP}=\mathrm{P}$ (try all possible log-long advice to search for a satisfying assignment in SAT problem).

Second, the situation for $\mathrm{BPP}$ is more complicated. Since $\mathrm{BPP}$ languages do not have short certificate, one cannot hope to deduce that $\mathrm{BPP}=\mathrm{P}$ from $\mathsf{BPP\subset P/}log$.

What we can deduce from $\mathsf{BPP\subset P/}log$ is that $\mathsf{BPP\cap NP=BPP\cap co}$-$\mathsf{NP=RP=co}$-$\mathsf{RP=P}$. That is, the part of $\mathsf{BPP}$ inside $\mathsf{NP\cup co}$-$\mathrm{NP}$ is just $\mathrm{P}$.