Answer
Because we can write:
$$\det(PAP^{-1})=\det(PP^{-1}A)=\det(A)$$
Work Step by Step
We make use of theorem 6, which holds because the are bothe square matrices, and commutativity of real and complex numbers:
$$\det(PAP^{-1})=\det(P)\det(A)\det(P^{-1})=\det(P)\det(P^{-1})\det(A)\\=\det(PP^{-1}A)=\det(IA)=\det(A)$$
We have now reached our conlussion $\det(PAP^{-1})=\det(A)$.
