The proposition is as follows :
Let $A$ be integrally closed in its quotient field $K$, and $B$ be integral closure of $A$ in finite Galois extension $L$ of $B$, with group $G$. Let $\mathfrak p $ be maximal ideal of $A$, $\mathfrak q$ be maximal ideal of B lying above $\mathfrak p $. Then $B/\mathfrak{q} $ is normal over $A/\mathfrak p$, and mapping $\sigma \mapsto \bar{\sigma} $ is homomorphism of $G_{\mathfrak q}$ (decomposition group of $\mathfrak q $) onto the Galois group of $B/\mathfrak q $ over $A/\mathfrak p $.
I have no problem with the part proving $\bar{B}/\bar{A} $ is normal extension. As the automorphism group is referred to as 'Galois group of $\bar{B}/\bar{A} $', I assume the proposition means that $\bar{B}/\bar{A} $ is normal and separable extension, but it seems that the separability part of the proof doesn't appear in the text.
I find it hard to prove the separability myself and actually it seems that it is not generally separable.
The separability does not always hold. It is true, however, if the residue field $A/\mathfrak{p}$ is finite.