By definition $P$ is a real symplectic matrix and your question is called the polar decomposition for symplectic matrices.
Let $Q=\sqrt{PP^t}$. Then the matrix $Q$ is symmetric and positive definite. Now take $S=Q^{-1}P$ and check that $SS^t=I$, that is, $S$ is orthogonal. Moreover, this decomposition is unique: if $P=Q_1S_1=Q_2S_2$, then $Q_1^2=Q_2S_2S_1^tQ_1=Q_2^2$ and thus, by the uniqueness of the square root for positive symmetric matrices, we get $Q_1=Q_2$ and then $S_1=S_2$. (Nothing new so far: this is the polar decomposition of any invertible matrix.)
From $P^tJP=J$ we get $P=J^{-1}(P^t)^{-1}J$. By using that $P=QS$ we obtain $P=J^{-1}(S^tQ)^{-1}J=(J^{-1}Q^{-1}J)(J^{-1}(S^t)^{-1}J).$ The matrix $J^{-1}Q^{-1}J$ is symmetric and positive definite, while $J^{-1}(S^t)^{-1}J$ is orthogonal. By the uniqueness of the polar decompositon we obtain that $Q=J^{-1}Q^{-1}J$ and $S=J^{-1}(S^t)^{-1}J$, therefore $Q$ and $S$ are also symplectic.