We consider a small category $\mathcal{C}$ with binary products, and we consider, for any objects $A, B$ of $\mathcal{C}$, the assignments
\begin{eqnarray*} F :\,\, &\mathcal{C}^\text{op} &\longrightarrow \textbf{Set} \\ &X &\mapsto \text{Hom}_{\mathcal{C}}(X \times A, B) \\ &X'\xrightarrow{\bar{f}} X &\mapsto - \circ f\times \text{id}_A. \end{eqnarray*}
After some fidgeting with compositions and reversals of the order of morphisms in opposite categories, I managed to prove this is a functor, and therefore, in fact, a presheaf (on $\mathcal{C}$). Now I am asked the following somewhat vague question: 'what does it say about $\mathcal{C}$ if this functor is representable?
In the preceding exercises, I have shown that $\mathcal{C}$ has binary products iff (for $y$ the Yoneda embedding) $y(A) \times y(B)$ is representable, and also that $\mathcal{C}$ has a terminal object iff a (c.q. 'the') terminal object of $[\mathcal{C}^\text{op}, \textbf{Set}]$ is representable. However I'm not sure how any it this (certainly the latter) ties into this, nor what the question is getting at. Can someone point me in a likely sensible direction?
Cheers.
Solution attempt:
Suppose $F$ is representable: $(\exists D \in \mathcal{C}_0)(F \cong y(D))$. Then we get, for any object $C$ of $\mathcal{C}$:
$$y(D)(C) = \text{Hom}(C, D) \leftrightarrow \text{Hom}(C \times A, B) = F(C). $$
If we furthermore assume that $B^A$ exists (with some map ev$_A : B^A \times A \to B)$, then we also have a bijection $$\text{Hom}(C \times A, B) \leftrightarrow \text{Hom}(C, B^A). $$ Combining this, this yields $$y(D)(C) \leftrightarrow \text{Hom}(C, B^A) = y(B^A)(C), $$ whence ($y$ being fully faithful and injective on objects) $D \cong B^A$.
Thus: if $(B^A, \text{ev}_A)$ exists, then $B^A \cong D$.
However, there is (in my mind) no reason to assume that $(B^A, \text{ev}_A)$ exists...
That aside: even if I were to prove that it does, i.e., that $B$ is exponentiable and $B^A \cong D$, what does that say about $\mathcal{C}$ as a whole? Is it that, as $A, B$ were arbitraty, $\mathcal{C}$ is ccc?