= Solution
The claim fails for a general extension. If $k$ is an infinite <field>, all the infinitely many maximal ideals $(x-a)$ of $k[x]$ contract to $(0)$ in $k$.
It still fails for an <integral extension>. Take a finite field $k=\mathbb F_q$ and
$$
B=\prod_{j\geq1}k.
$$
Every $b\in B$ satisfies the monic equation $b^q-b=0$, so $B$ is integral over the diagonal copy of $k$. The coordinate kernels are infinitely many distinct <maximal ideal>[maximal ideals], all lying over $(0)$.
The claim is true for a <module-finite ring extension>. The primes above $\mathfrak p$ correspond to the primes of the <fiber ring>
$$
B\otimes_A\kappa(\mathfrak p).
$$
This is a finite-dimensional algebra over the <residue field> $\kappa(\mathfrak p)$, hence an <Artinian ring>, and an Artinian ring has only finitely many prime ideals.
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