If G acts on a finite set S, then S can be written as a disjoint union (See Hungerford's algebra p 93)
$\displaystyle S=S_0 \cup \bar{x_1} \cup \bar{x_2} \cup \cdots \cup \bar{x_n}$, where $\displaystyle S_0=\{x \in S: g\cdot x=x \text{ for all } g \in G\}$, $\displaystyle |\bar{x_i}|=|G:G_{x_i}|$, and $\displaystyle |\bar{x_i}|>1$ for all i. We see that $\displaystyle |\bar{x_i}|$ is divisible by p by Lagrange's theorem, since a p-group G acts on S. By assumption, S is not divisible by p, so $\displaystyle S_0$ is not empty and $\displaystyle p \nmid|S_0|$. An orbit of $\displaystyle x \in S$, denoted $\displaystyle \bar{x}$, has a single element iff $\displaystyle x \in S_0$. Can you conclude it from here?