= Solution
Form the finite-dimensional $k$-algebra
$$
C=B\otimes_{A,f}k.
$$
Because a finite extension is integral, the <Lying-over theorem> supplies a prime of $B$ above $\ker f$; after localization and extension of the residue field to $k$, this shows $C\ne0$. Therefore $C$ has at least one <maximal ideal>.
As an <Artinian ring>, $C$ has only finitely many maximal ideals. For each such ideal $\mathfrak n$, the quotient $C/\mathfrak n$ is a finite <field extension> of the algebraically closed field $k$, so it equals $k$. Consequently the quotient maps $C\to k$ are in bijection with the required extensions $g:B\to k$. The set of extensions is therefore finite and nonempty.
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