Looking for elementary proof for $\rm~det(\textbf{P})$ = $\pm 1$ if $P$ is an orthogonal matrix. I prefer a proof without using determinant of transpose matrix.
My First Proof, with $\det(\textbf{P}^{t}) = \det(\textbf{P})$
If $P$ is orthogonal matrix, $\textbf{P}^{t}=\textbf{P}^{-1}$. So,
$$\det(\textbf{P}^{t}\textbf{P})=\det(\textbf{I}) \implies \det(\textbf{P}^{t}\textbf{P}) = 1 \implies \det(\textbf{P}^{t}) \det(\textbf{P}) = 1$$
because $ \det(\textbf{P}^{t}) = \det(\textbf{P})$.
Therefore, $\det(\textbf{P}^{t})=\det(\textbf{P}) = \pm 1$.
My Second Proof, without $\det(\textbf{P}^{t}) = \det(\textbf{P})$
Let $\lambda$ be an eigenvalue for orthogonal matrix $\textbf{P}$.
$$\textbf{P}\vec{v}=\lambda\vec{v},\quad\vec{v} \neq \vec{0}$$
$$\implies \rVert \textbf{P}\vec{v}\lVert = \lVert \lambda\vec{v} \lVert \implies \rVert \textbf{P}\vec{v} \lVert = \rvert \lambda \ \lvert \lVert \vec{v} \rVert = 1$$
because $\lVert \vec{v} \rVert = 1$
Therefore $\lvert \lambda \rvert = 1$
because $\textbf{P}$ is orthonormal matrix.
$$\textbf{P} = \textbf{P}^{t}$$
Therefore $\textbf{P} = \textbf{U}^{t}\Lambda\textbf{U} $
$$\det(\textbf{U}^{t}\Lambda\textbf{U}) = \det(\textbf{U}^{t})\det(\Lambda)\det(\textbf{U})$$
$$\det(\textbf{U}^{t}\Lambda\textbf{U}) = \det(\textbf{U}^{t})\det(\textbf{U})\det(\Lambda)$$
$$\det(\textbf{U}^{t})\det(\textbf{U}) = 1 $$
(because $\textbf{U}$ is orthogonal matrix})
$$\det(\textbf{P}) = \det(\textbf{U}^{t}\Lambda\textbf{U}) = \det(\Lambda) = \prod_{i=1}^{n}\lambda_i = \pm 1 $$
$$\implies \det(\textbf{P}) = \pm 1$$
Best Answer
Here is a geometric argument that can be used for real orthogonal matrices: If $T:\ V\to V$ is an arbitrary linear transformation of a finite dimensional euclidean vector space $V$ then the volumes of arbitrary measurable sets $A\subset V$, in particular of balls or cubes, are multiplied by $\bigl|\det(T)\bigr|$. That is to say, one has $${\rm vol}\bigl(T(A)\bigr)=\bigl|\det(T)\bigr|\ {\rm vol}(A)\ .$$ Now an orthogonal transformation $T$ transforms the unit ball $B:=\bigl\{x\in{\mathbb R}^n\bigm| |x|\leq1\bigr\}$ onto itself, and ${\rm vol}(B)>0$. It follows that $|\det(T)\bigr|=1$.