Given the matrix equation $PD=C^TF^T$ where $P$ is positive definite and $F$ can be set arbitrarily. Matrices $C$ and $D$ are of full rank. Prove that $\operatorname{rank}(CD) = \operatorname{rank}(D)$ is a sufficient existence condition for the solvability of the matrix equation above.
So far I have only been able to prove that $\operatorname{rank}(CD) = \operatorname{rank}(D)$ is a necessary existence condition. This problem is related to a field in control theory. The above condition represents one of the existence conditions of a so called sliding mode observer.
There are still some things about your question which seem odd.
If $C$ is an $n\times m$ matrix, then $D$ has to be $m \times k$ for some $k$ in order for $CD$ to be defined. Saying that $C$ and $D$ are full-rank means $\operatorname{rank}(C) = n$ and $\operatorname{rank}(D) = m$.
But $C$ being full-rank also means that for any $n$-vector $w$, there must be an $m$-vector $v$ such that that $Cv = w$. And $D$ being full-rank means there must be a $k$-vector $u$ such that $Du = v$ and so $CDu = w$. Since $CD$ is surjective, it is also full-rank, and hence $\operatorname{rank}(CD) = n$. Therefore your condition that $\operatorname{rank}(CD) = \operatorname{rank}(D)$ requires $m = n$ and therefore $C$ is a square matrix.
As a full-rank square matrix, $C$ is invertible. Thus $$F = D^TP^TC^{-1}$$ is your solution.
The only thing the rank condition is required for here is to show that $C$ is a square matrix. And in fact, when $C$ is a full-rank square matrix, and $D$ is also full-rank matrix, $\operatorname{rank}(CD) = \operatorname{rank}(D)$ must be true.
Since we generally know what the dimensions of the matrices we are working with are, $\operatorname{rank}(CD) = \operatorname{rank}(D)$ seems like an awfully convoluted way of checking that $C$ is square.