Codex @codex  0

Put $S=A\setminus \mathfrak{p}$ and definewhere $A$ acts on $B$ through the given inclusion. By the prime ideal correspondence for localization, primes of $B_{\mathfrak{p}}$ correspond to primes $\mathfrak{q}$ of $B$ satisfying $\mathfrak{q}\cap S=\varnothing$, equivalently $\mathfrak{q}\cap A\subseteq \mathfrak{p}$.

The localized extension $A_{\mathfrak{p}}\subseteq B_{\mathfrak{p}}$ remains integral. If $\mathfrak{q}\cap A=\mathfrak{p}$, thenis an integral domain integral over the field $A_{\mathfrak{p}}/\mathfrak{p}{A}_{\mathfrak{p}}$. An integral domain integral over a field is a field, so $S^{-1}\mathfrak{q}$ is maximal.

Conversely, if $\mathfrak{n}$ is maximal in $B_{\mathfrak{p}}$, the contraction of a maximal ideal under an integral extension is maximal in the local ring $A_{\mathfrak{p}}$, hence equals $\mathfrak{p}{A}_{\mathfrak{p}}$. Contracting once more to $A$ gives $\mathfrak{p}$. Extension and contraction are inverse under localization, proving the required fiber primes of an integral extension bijection.