A more general version of the result Beenakker cites that allows for normal $B$ (not just symmetric):$\newcommand\tr{\operatorname{tr}}\newcommand\E{\mathbb{E}}$
Take $x\sim\mathcal{N}(0,I)$ and recall that $\tr(C) = \E[x^\mathsf{T}Cx]$. If $B = Q \Lambda Q^*$ is unitarily diagonalizable let $\Lambda^{1/2}$ be a (possibly complex) square root of $\Lambda$. Then using the cyclic property of the trace $$ \begin{align} \tr(AB) &= \tr(\Lambda^{1/2} Q^*AQ\Lambda^{1/2})\\ &= \E[(\Lambda^{1/2}x)^\mathsf{T} Q^*AQ (\Lambda^{1/2}x)]\\ &= \sum_{i\neq j} \lambda_i^{1/2}\lambda_j^{1/2}\E[x_ix_j](Q^*AQ)_{ij} + \sum_{i} \lambda_i\E[x_i^2](Q^*AQ)_{ii}\\ &= \sum_{i} \lambda_i(Q^*AQ)_{ii}. \end{align} $$ From here, since $(Q^*AQ)_{ii}\geq 0$ by the PSD condition, it follows that $$\tr(AB) \leq \lambda_{\max}\sum_{i} (P^{-1}A P)_{ii} = \lambda_{\max}\tr(P^{-1}AP) = \lambda_{\max}\tr(A)$$ and $$\tr(AB) \geq \lambda_{\min}\sum_{i} (P^{-1}A P)_{ii} = \lambda_{\min}\tr(P^{-1}AP) = \lambda_{\min}\tr(A)$$ as desired.