Let G be a group of order 24 that is not isomorphic to S4. Then one of its Sylow subgroups is normal.

It has to do with the fact that every (non-empty) fiber of a homomorphism is a coset of the kernel. That is, if $\varphi:G\to H$ is a homomorphism, and $h\in\operatorname{im}\varphi,$ then the fiber of $h$ under $\varphi$ is the set $$\{g\in G:\varphi(g)=h\},$$ and is a coset of $\ker\varphi$ in $G$. I outline the proof of this fact (from a linear algebra standpoint) in my answer here, and not much changes in the more general case.

Since $\ker f$ has order two, then for any $\sigma\in S_4,$ we have $f^{-1}(\sigma)$ has cardinality either $2$ or $0$. Since we're assuming that $A_4=\operatorname{im}f,$ then for each $\sigma\in A_4$ (and in particular for each $\sigma\in P_2$) we have $f^{-1}(\sigma)$ has cardinality $2$. Since $P_2$ has $4$ elements by the referenced exercise, then $f^{-1}(P_2)$ is a union of $4$ pairwise disjoint sets of cardinality $2$, meaning that $f^{-1}(P_2)$ has order $8$.

Does that help?