$• \ \text{Alternativa (A):}$ $\color{orangered}{\text{Incorreta}}$ Se $AB$ é simétrica, então temos: $AB = (AB)^t$, vejamos: \begin{matrix} AB = AP^tAP &,& (AB)^t = (AP^tAP)^t = PAP^tA \end{matrix} \begin{matrix} AP^tAP \nRightarrow PAP^tA &\Rightarrow& AB \ne (AB)^t \end{matrix} $• \ \text{Alternativa (B):}$ $\color{orangered}{\text{Incorreta}}$ \begin{matrix} BA= P^tAPA &,& (BA)^t = (P^tAPA)^t = APAP^t \end{matrix} \begin{matrix} P^tAPA \nRightarrow APAP^t &\Rightarrow& BA \ne (BA)^t \end{matrix} $• \ \text{Alternativa (C):}$ $\color{royalblue}{\text{Correta}}$ \begin{matrix} det(B) = det(P^t) \ . \ det(A) \ . \ det(P) \\ \\ \color{gray}{det(P) \ . \ det(P^t) = det(I) = 1} \\ \\ det(B) = det(A) \end{matrix} $• \ \text{Alternativa (D):}$ $\color{orangered}{\text{Incorreta}}$ \begin{matrix} AB = AP^tAP &,& BA= P^tAPA \end{matrix} \begin{matrix}BA \ne AB \end{matrix} $• \ \text{Alternativa (E):}$ $\color{orangered}{\text{Incorreta}}$ \begin{matrix} B = P^tAP &,& B^t= PP^tA = IA = A \\ \\ BB^t = P^tAPA && B^tB = APAP^t \end{matrix} \begin{matrix}BB^t \ne B^tB \ne I \end{matrix}