= Solution
By part (b), $i:P\to H_B$ is the <equaliser> of $ip$ and $1_{H_B}$. Full faithfulness of the <Yoneda embedding> writes $ip=H_b$ for a unique <endomorphism> $b:B\to B$. Choose an <equaliser> $e:A\to B$ of $b$ and $1_B$ in $\mathcal C$.
For each $c$, the maps $c\to A$ are precisely the maps $u:c\to B$ with $bu=u$. Thus $H_e:H_A\to H_B$ is the pointwise <equaliser> of $H_b$ and $1_{H_B}$, hence their <equaliser> in the <presheaf category>. Both $i$ and $H_e$ have the same <universal property>, so their vertices are isomorphic. This proves the <representable retract criterion>:
$$
\boxed{P\cong H_A.}
$$
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