Let $G$ be a finite group of order $n$ and $p$ a prime number. Write $n=p^rm$ for some $r\in\mathbb{N}$ and $m\in\mathbb{N}_{\geq1}$ such that $m\not\in p\mathbb{Z}$. Let $$E=\{X\subset G\ :\ |X|=p^r\}.$$ Then $|E|=\binom{n}{p^r}$; since $\binom{n}{p^r}\not\equiv0\pmod{p}$, it follows that $|E|\not\equiv0\pmod{p}$.
Let $G$ act on itself by left translation: i.e. the map $\gamma:G\rightarrow\mathfrak{S}_G,\,g\mapsto(h\mapsto gh)$, is a homomorphism. Extend this action canonically to $\mathfrak{P}(G)$. Since $E$ is stable under this action, the map $$\varphi:G\rightarrow\mathfrak{S}_E,\,g\mapsto(X\mapsto gX),$$ is a homomorphism.
There exists $X\in E$ whose orbit has non-zero cardinal mod $p$.
Attempt:
$p^r\leq n$ implies that $\binom{n}{p^r}\ne0$; hence, $E\ne\emptyset$. Let $X\in E$. Then $X\subset G$ and $|X|=p^r$. Obviously $\text{orb}(X)$ is nonempty. But why should it be the case that $$|\text{orb}(X)|\not\equiv 0\ \pmod{p}?$$