This question is on the hint that the book gives to finish the exercise.
Namely, if $f: A \rightarrow B$ a homomorphism of rings, $q$ a prime ideal of $B$ and $p = q^c$, then $B_q$ is a local ring of $B_p$.
I can't make sense out of this statement... Is that saying the image of $q$ of in $B_p$ a prime ideal of $p$ through the map $b \mapsto \frac{b}{1}$? But this map is not always injective. But otherwise how do we understand $B_q$ as a localization of $B_p$ over some prime ideal? I don't understand...
Can someone give an explicit explanation of this statement?
I think what's confusing is first what $B_\mathfrak{p}$ is.
$B_\mathfrak{p}$ is the localization of $B$ at $\mathfrak{p}$, considered as an $A$-module via the morphism $f\colon A \to B$, i.e., $B_\mathfrak{p} \cong B \otimes_A A_\mathfrak{p}$, and in particular it is defined by localizing with respect to the multiplicative set $\{f(a) \mid a \in A \setminus \mathfrak{p}\}$ (it is multiplicative since $f$ is a ring homomorphism).
Then, by "$B_\mathfrak{q}$ is a local ring of $B_\mathfrak{p}$" is that $B_\mathfrak{q}$ is obtained by further localizing with respect to a prime ideal in $B$, and we can see this by the fact that $\{f(a) \mid a \in A \setminus \mathfrak{p}\} \subseteq \{b \mid b \in B \setminus \mathfrak{q}\}$.
One thing to mention is that $B_\mathfrak{p}$ is not necessarily a local ring: if you trust Exercise $3.21$, $\operatorname{Spec} B_\mathfrak{p}$ corresponds to the preimage of primes contained in $\mathfrak{p}$ under the map $f^*\colon \operatorname{Spec} B \to \operatorname{Spec} A$, and $\operatorname{Spec} B_\mathfrak{p}$ can be large in general.
Now the conclusion of the problem follows by property (iii) in Exercise 3.16.