= Solution
Choose algebra generators $b_1,\ldots,b_n$ for $B$ over $A$. Because the extension is integral, each $b_i$ satisfies a monic equation
$$
b_i^{e_i}+a_{i,e_i-1}b_i^{e_i-1}+\cdots+a_{i,0}=0\qquad(a_{i,j}\in A).
$$
Repeatedly reduce any exponent at least $e_i$ using this equation. Since the generators commute, every polynomial in them becomes an $A$-linear combination of the finitely many monomials
$$
b_1^{\nu_1}\cdots b_n^{\nu_n},\qquad0\leq\nu_i<e_i.
$$
Thus $B$ is a <finitely generated module> over $A$, with at most $D=\prod_i e_i$ generators. This proves the relevant <subalgebra generated by finitely many integral elements> assertion rather than merely invoking it.
For a fixed <prime ideal> $\mathfrak p\subset A$, form the <fiber ring> $C=B\otimes_A\kappa(\mathfrak p)$. Tensoring the finite generating list shows $\dim_{\kappa(\mathfrak p)}C\leq D$: its <pure tensors> with one span the fiber over the <residue field> $\kappa(\mathfrak p)$. Part (b) therefore gives only finitely many primes in $C$. By the canonical bijection in part (a), these are precisely the primes of $B$ contracting to $\mathfrak p$. Hence
$$
\boxed{\#\{\mathfrak q\in\operatorname{Spec}B:A\cap\mathfrak q=\mathfrak p\}\leq\dim_{\kappa(\mathfrak p)}(B\otimes_A\kappa(\mathfrak p))\leq D<\infty.}
$$
No <Noetherian> hypothesis on $A$ is needed. For an empty algebra-generating list, $B=A$ and the empty product $D=1$ gives the same conclusion.
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