I am considering the previous version of your question which contained two parts.
Part 1) We assume $Q<P$ which means $P-Q$ is a strictly positive matrix. Now one can show that if $X$ is a solution of the equation in your question then $A={(P-Q)}^{-1/2}XP^{1/2}$ satisfies $AA^T=Id$
Part 2) The answer is yes if we further assume that $PQ=QP$.
Note that the spectrum of the solution $X$ is equal to the spectrum of $T{(P-Q)}^{1/2}P^{-1/2}$. Now the commutativity assumption $PQ=QP$ implies that $H={(P-Q)}^{1/2}P^{-1/2}$ is a positive matrix.
Now we use the following lemma:
Lemma: If $T$ is an orthonormal matrix and $H$ is a positive matrix Then $\lambda_{min} (TH)= \lambda_{min}(H)$ and $\lambda_{Max} (TH)=\lambda_{Max}(H)$.
Proof: It is sufficient to prove the following statement. Then the lemma is a consequence of a simple rescaling and consideration of the fact that "The spectrum of the inverse is equal to the inverse of the spectrum"
Statement: If all eigenvaluse of a positive matrix $H$ lie in the interior of the unit disc of the complex plane then the same is true for all eigenvalues of $TH$ where $T$ is a unitary matrix:
Proof of the Statement: WLOG we may assume that $H$ is a diagonal matrix because every positive matrix is unitary equivalent to a diagonal matrix. For diagonal matrices the statement is obvious. Because if $THV=\lambda V$ for some $\lambda \in \mathbb{C}$ with $|\lambda|>1$ and $V \in \mathbb{C}^n$ then $|HV|>|V|$ which is impossible since all entries of diagonal matrix $H$ are positive element less than $1$.