I am currently reading "Character Theory of Finite Groups". If the Sylow $p$-subgroup $P$ is abelian, then by Theorem $3.13, p$ is the exact power of $p$ diving $|G: Z(G)| = |G|$ since $G$ is simple.
I don’t know how to prove the hint.
If $\chi$ restricted to $P$ is irreducible, then corollary $2.28$ proves the hint, but I don’t know how to prove $\chi$ is irreducible when restricted to $P$, or if that’s the right idea? I know since $Z(P)$ is not trivial, if $Z(P) \leq Z(\chi)$, then since $G$ is simple, we would have $G = Z(\chi)$, but not sure how that helps?
Any reference here is to Isaac's book Character Theory of Finite Groups.
Some preparatory remarks. Let $P \in Syl_p(G)$ ($P$ exists and is non-trivial, since $\chi(1)=p \mid |G|$). Now $\chi_P=\sum a_{\theta}\theta$ for integers $a_{\theta} \gt 0$ and $\theta \in Irr(P)$. Since $\theta(1) \mid |P|$, each $\theta(1)$ is a power of $p$. Clearly, $\chi_P(1)=\chi(1)=p=\sum a_{\theta}\theta(1) \geq \theta(1)$ for each of these irreducible constituents $\theta$ of $\chi_P$. Hence, $\theta(1) \in \{1,p\}$. Further, observe that $\chi$ is non-linear, so $G$ cannot be abelian. Since $G$ is simple, $Z(G)=1$, and the non-linearity of $\chi$ gives $Z(\chi) \lt G$, whence $Z(\chi)=1$ and so is $ker(\chi)=1$ ($\chi$ is faithful). Now we have two cases.
(1) If $\theta(1)=p$ for some $\theta$, then $\chi_P=\theta$ (of course, the corresponding $a_{\theta}=1$). Hence $\chi_P$ is irreducible. But then $ker(\chi_P)=P \cap ker(\chi)=1$, so (Lemma 2.27(f)) $Z(P)\subseteq Z(\chi_P)=P \cap Z(\chi)=1$, contradicting $P$ being a non-trivial $p$-group. Hence:
(2) $\theta(1)=1$ for all irreducible constituents $\theta$ of $\chi_P$, that is, $\chi_P$ is the sum of some linear characters. But then, (Lemma (2.21) and (2.23)(a)), $1=ker(\chi_P)=\bigcap_{\theta} ker(\theta) \supseteq P'$. We conclude that $P$ is abelian.
Note that we have proved sofar that if $G$ is a simple group having an irreducible character of degree $p$, then a $p$-subgroup must be abelian! The proof can now be finished by applying Theorem(3.13) and it is here that we are using that $P$ is a Sylow $p$-subgroup.
Note A similar line of reasoning as above gives the following. Let $G$ be a simple group and suppose $\chi\in Irr(G)$ with $\chi(1)=p$ a prime. Let $q$ be another prime, with $q \mid |G|$ and $q \gt p$. Then a $q$-subgroup of $G$ is abelian. In particular a Sylow $q$-subgroup is abelian.