 Solution

If $Av=\lambda v$, then $p(A)v=p(\lambda )v$, proving one inclusion. Conversely, if $\mu$ is an eigenvalue of $p(A)$, factor $p(z)-\mu =c{\prod }_j(z-{\lambda }_j)$. If none of the ${\lambda }_j$ were eigenvalues of $A$, every $A-{\lambda }_{j}I$ would be invertible, making $p(A)-\mu I$ invertible, a contradiction. Thus $\mu =p(\lambda )$ for some eigenvalue $\lambda$ of $A$.