Answer
See below
Work Step by Step
According to property P9:
$\det(S^{-1}AS)^2\\=\det(S^{-1}AS)(S^{-1}AS)\\=\det(S^{-1}AS)\det(S^{-1}AS)\\=\det(S^{-1})\det(A)\det(S)\det(S^{-1})\det(A)\det(S)\\=\det(S^{-1})\det(S)\det(S^{-1})\det(S)\det(A)\det(A)$
Since $A$ and $S$ are $n \times n$ matrices with $S$ invertible, we have:
$\det(S^{-1})\det(S)\det(S^{-1})\det(S)\det(A)\det(A)\\
=\det(S^{-1}S)\det(S^{-1}S)(\det(A))^2\\
=\det(I_n)\det(I_n)(\det(A))^2\\
=(\det(A))^2$
Consequently, $\det(S^{-1}AS)^2=(\det(A))^2$
