 Solution

The degree of a chromatic polynomial is the number of vertices and the negative of its next coefficient is the number of edges. Equality with $p_{K_n}$ therefore forces $n$ vertices and $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ edges. A simple graph on those vertices with that many edges contains every possible edge, so $\boxed{G\cong K_n}$.