The above paragraph claims that $C=p^{-1}(D)$ for some $D\subset Y$. I think $D=\{y\in Y| p^{-1}(y)\cap C\ne \emptyset\}$. Is this correct? Even if so, I can't prove it.
Is the converse true? That is, if $C$ is saturated, then $C=p^{-1}(D)$. I think I managed to prove one inclusion. If $y\in D$, then I must prove that $p^{-1}(y)\in C$. Let $x\in p^{-1}(y)$. Then $p(x)=y\in D$ and so $x=p^{-1}(y)$. Since $y\in D$, $x\cap C\ne \emptyset$, and since $C$ is saturated, $x\in C$. This proves $p^{-1}(D)\subset C$
Suppose that $C$ is saturated. I claim that $p^{-1}[p[C]] = C$ and so $C$ is the pre-image of a subset of $Y$, as claimed (so we can take $D = p[C]$).
Note that by definition we always have $p^{-1}[p[C]] \supseteq C$, because $c \in C$ implies $p(c) \in p[C]$ and so $c \in p^{-1}[p[C]]$. Now suppose that $x \in p^{-1}[p[C]]$. This means that $p(x) \in p[C]$ or there is a $c \in C$ such that $p(c) = p(x)$. So $C$ intersects $p^{-1}[\{p(x)\}$ and by being saturated, $p^{-1}[\{p(x)\}] \subseteq C$ ($C$ contains every fibre it intersects). And in particular, $x \in p^{-1}[\{p(x)\}] \subseteq C$, showing the reverse inclusion.
Reversely it's easy to check that $C = p^{-1}[D]$ is indeed always a saturated subset of $X$.