= Solution
{c}
Write $h_A=\mathcal C(A,-)$. The covariant <Yoneda lemma> gives the natural bijection
$$
\operatorname{Nat}(h_A,F)\cong F(A),\qquad\alpha\longmapsto\alpha_A(1_A),
$$
whose inverse sends $x\in F(A)$ to the <natural transformation> with component $u:A\to B\mapsto F(u)(x)$. This also identifies the two variables' naturality.
To prove that $h_A$ is a <projective object in a category>, let $p:G\twoheadrightarrow F$ be an <epimorphism> in the <functor category> and $\alpha:h_A\to F$. By <pointwise epimorphism in a functor category>, lift $\alpha_A(1_A)$ through the <surjective function> $p_A$. The <Yoneda lemma> turns this lift into $\beta:h_A\to G$ with $p\beta=\alpha$.
For a <small category>, the set of pairs $(A,x)$ with $x\in F(A)$ indexes the canonical map
$$
p:\coprod_{(A,x)}h_A\longrightarrow F,\qquad
p_B(A,x,u)=F(u)(x).
$$
It is natural, and every $y\in F(B)$ is the image of $(B,y,1_B)$. Thus it is an <epimorphism>, establishing the <projective cover of a set-valued functor by representables> construction. Here cover means a projective presentation; no minimality is asserted.
For the equivalence's first step, if $f:B\to D$ is a <monomorphism>, then $h_A(f)$, the map $u\mapsto fu$, is injective for every $A$. Hence every <representable functor> is a <monofunctor>. Conversely, injectivity of every $h_A(f)$ is exactly left cancellation against every $u,v:A\to B$, so it implies that $f$ is a <monomorphism>. Therefore
$$
\boxed{\text{(a)}\Longleftrightarrow\text{(b)},\qquad
\coprod_{(A,x)}h_A\twoheadrightarrow F.}
$$
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