Prove that $\left|\begin{smallmatrix}a&-b&-c&-d\\b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{smallmatrix}\right|=(a^2+b^2+c^2+d^2)^2.$

Calculate $$ P P^T $$ then think about it.

Or $$ P^T P $$

Generally， if $B$ is symmetric such that $A^TB=BA$, then $$ \det \begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix} = \det (AA^T +BB^T). $$ In fact $$ \begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}\begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}^T=\begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}\begin{bmatrix} A^T & -B^T \\ B^T & A^T \\ \end{bmatrix} = \begin{bmatrix} AA^T+BB^T & -AB^T+BA^T \\ -BA^T+AB^T & AA^T+BB^T \\ \end{bmatrix}=\begin{bmatrix} AA^T+BB^T & 0 \\ 0 & AA^T+BB^T \\ \end{bmatrix} $$ and hence $$ \det\begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}=\sqrt{\det\begin{bmatrix} AA^T+BB^T & 0 \\ 0 & AA^T+BB^T \\ \end{bmatrix}}=\det(AA^T+BB^T).$$