You can suppose that $A$ is not only a local ring, but actually a field. Let $f \colon A \to B$ be a finite ring homomorphism, let $\mathfrak{p}$ be a prime ideal of $A$, let $f^* \colon \mathrm{Spec} B \to \mathrm{Spec} A$ the associated map, i.e. $f^*(\mathfrak{q}) = f^{-1}(\mathfrak{q})$ for every prime ideal $\mathfrak{q}$ of $B$.
(1) You can suppose that $A$ is local and $\mathfrak{p}$ is the unique maximal ideal of $A$. Consider $S = A \setminus \mathfrak{p}$. $S^{-1}f \colon A_\mathfrak{p} \to S^{-1}B$ is a finite homomorphism of rings. You should check that $(f^*)^{-1}(\mathfrak{p})$ is in bijection with $((S^{-1}f)^*)^{-1}(\mathfrak{p}A_{\mathfrak{p}})$. Hence you can replace $A$ and $\mathfrak p$ with $A_\mathfrak{p}$ and $\mathfrak{p} A_\mathfrak{p}$.
(2) You can suppose that $A$ is a field and $\mathfrak p = 0$. $f$ induces a finite homomorphism $\bar{f} \colon A/\mathfrak{p} \to B/\mathfrak{p}B$. Check that the fiber of $f^*$ over $\mathfrak{p}$ is in bijection with the fiber of $\bar{f}^*$ over $\overline{\mathfrak p} = 0$. You can replace $A$ with the field $A/\mathfrak{p}$ and $B$ with $B/\mathfrak{p}B$.
Steps (1) and (2) can be unified in a unique step: replace $A$ with $k(\mathfrak{p})$, and $B$ with $B \otimes_A k(\mathfrak{p})$, where $k(\mathfrak{p})$ is the residue field of $\mathfrak{p}$. This is described in Proposition 3.1.16 of Liu's Algebraic geometry and arithmetic curves.
The proof continues in my answer here.