This is assuming $Cat$ is the category of small categories, otherwise $GC$ (for $C$ in $Cat$) would not be a set and can thus not be a preorder.
You can define the map $\phi: \hom(GC, D) \to \hom(C, FD)$ as follows. Given a morphism of preorders $f: GC \to D$, we define a functor $g: C \to FD$ by letting $g$ be the same as $f$ on objects (since the objects in $C$ and $FD$ are the elements in $GC$ and $D$ respectively). For any arrow $X \to Y$ in $C$, we have that $X \leq Y$ in $GC$, so since $f$ is a morphism of preorders we have $f(X) \leq f(Y)$ in $D$. That means that we have exactly one arrow $g(X) \to g(Y)$ in $FD$, so we send the arrow $X \to Y$ there. It should be clear that this defines an actual functor $g: C \to FD$.
The operation $\phi$ has an inverse, by taking a functor $g: C \to FD$ and defining a morphism of preorders $f: GC \to D$ by letting $f$ be setting $f(X) = g(X)$ for all elements in $GC$ (again, which were objects in $C$). The fact that $g$ was a functor now guarantees that $f$ is a morphism of preorders.
Essentially this is saying that there is a unique way to extend a morphism of preorders to a functor of categories, because there is exactly one place we can send the arrows.
Of course, you would still have to check that $\phi$ is a natural bijection, but that is not hard and I leave that as an exercise to the reader.