Assume $\,|G|=p^nm\,,\,(p,m)=1$:
I assume you already know the following Sylow theorems: Sylow $p$-sbgps. exist in $\,G\,$ , they all are conjugate in $\,G\,$ and their number equals $\,1\pmod p:$ let $\,K\,$ be a $p$-sbgp. and let $\,\mathcal{P}:=\{P\leq G \mid P\text{ is a Sylow }p\text{-sbgp. of }\, G\}\,$. As noted above, $\,|\mathcal{P}|\equiv 1\pmod p$.
Let now $K$ act on $\,\mathcal{P}\,$ by conjugation: $\,k\in K\,,\, P\in\mathcal{P}\Longrightarrow k\cdot P\to k^{-1}Pk=:P^k\,$ . If there is an orbit with one single element , say $\,\mathcal{O}_Q=\{Q\}\,\,,\,\,\text{for some}\,Q\in\mathcal{P}$, then $\,Q^k=Q\,\,\forall\,k\in K\Longrightarrow QK=KQ\Longrightarrow QK\,$ is a $p$-subgroup of $\,G\,$ , so if $\,K\,$ is not contained in any Sylow $p$-sbgp then $\,|QK|>|Q|=p^n\,$, which is absurd, so that all the orbit have size a power of $p$, but this means $\,|\mathcal{P}|\equiv 0\pmod p$, which of course is also absurd since, as we mentioned above, $\,|\mathcal{P}|\equiv1\pmod p\Longrightarrow \,$ it must be that $\,K\,$ is contained in some Sylow $p$-sbgp.