Integral extensions: one prime lying over implies equal localization

First of all, it should be pointed out that $B_{\mathfrak{p}}$ is the localization of the $A$ module $B$ at $\mathfrak{p}$. It is isomorphic to the ring $A_{\mathfrak{p}} \otimes_A\ B$ as an $A$ module. It need not be the case that $\mathfrak{p}$ is a prime ideal of $B$, thus to localize $B$ at $\mathfrak{p}$ as a ring doesn't make sense (for $B -\mathfrak{p}$ would not be a multiplicatively closed set). This should clear up your confusion about $B_{\mathfrak{p}}$ not a priori being a local ring.

That any maximal ideal $\mathfrak{m}$ of $B_{\mathfrak{p}}$ lies over $\mathfrak{p}A_{\mathfrak{p}}$ is a standard result used to prove the going up theorem. For since $B_{\mathfrak{p}}$ is integral over $A_{\mathfrak{p}}$, we then have that $B_{\mathfrak{p}} /\mathfrak{m}$ is integral over $A_{\mathfrak{p}}/(\mathfrak{m}\cap A_{\mathfrak{p}})$. But $B_{\mathfrak{p}} /\mathfrak{m}$ is a field, so $A_{\mathfrak{p}}/(\mathfrak{m}\cap A_{\mathfrak{p}})$ must be one as well, whence $\mathfrak{m}$ lies over $\mathfrak{p}A_{\mathfrak{p}}$ since this is the unique maximal ideal of $A_{\mathfrak{p}}$.