Theorem 5.10. (Third Sylow Theorem) If $G$ is a finite group and $p$ a prime, then the number of Sylow $p$-subgroups of $G$ divides $|G|$ and is of the form $kp+1$ for some $k\geq O$.
Proof: By the second Sylow Theorem the number of Sylow $p$-subgroups is the number of conjugates of any one of them, say $P$. But this number is $[G:N_G(P)]$, a divisor of $|G|$, by Corollary 4.4. Let $S$ be the set of all Sylow $p$-subgroups of $G$ and let $P$ act on $S$ by conjugation. Then $Q\in S_0$ if and only if $xQx^{1}= Q$ for all $x\in P$. The latter condition holds if and only if $P \lt N_G(Q)$. Both $P$ and $Q$ are Sylow $p$-subgroups of $G$ and hence of $N_G(Q)$ and are therefore conjugate in $N_G(Q)$. But since $Q$ is normal in $N_G(Q)$, this can only occur if $Q = P$. Therefore, $S_0 =\{P\}$ and by Lemma 5.1, $|S|=|S_0|= 1\pmod p$. Hence $|S|=kp+1$.
Question: I don’t understand “But since $Q$ is normal in $N_G(Q)$, this can only occur if $Q = P$” sentence. By second Sylow theorem, $\exists x\in N_G(Q)=\{g\in G\mid gQg^{-1}=Q\}$ such that $P=xQx^{-1}=Q$. We’re simply using definition of $N_G(P)$.
I think, Author should have added $|G|=p^nm$ where $(p,m)=1$ (or equivalently $p$ divides $|G|$) condition in the hypothesis to guarantee existence of Sylow $p$-subgroup of $G$.
Edit: I see why Author phrase that sentence in that particular way. If $P$ is Sylow $p$-subgroup of $G$ and normal in $G$, then $P$ is the only Sylow $p$-subgroup of $G$. It’s proof follows directly from second Sylow theorem.