Well, that's equivalent to C-1PC=Q, isn't it?
solution:
What you want to do is to show, that they have the same char polynomial (up to scalarmultiplication). But that's easy since det(\mathbf{P} -\lambda \mathbf{I})=det(\mathbf{C^{-1}QC} -\lambda \mathbf{C^{-1}C})=det(\mathbf{C^{-1}}(\mathbf{Q}...