Let $B$ be a complete boolian algebra, then every object of the topos $Sh(B)$ of sheaves on $B$ is projective.
This is exercise 4.15.d of the book sheaves in geometry and logic.
Do I have to show that every object of the topos $Sh(B)$ is a retract of a sum of representatives?
On
You can actually find the solution later in the same book. Proposition VI.1.8 states that all epimorphisms split in a topos $T$ if it is generated by subobjects of $1$ and if for all $E \in T,$ $Sub(E)$ is a complete Boolean algebra. The proof is essentially what Daniel Schepler provided here: the maximal subobject for which a section of a given epi exists has to be the epi's codomain itself.
As in (i) of your exercise, $A \twoheadrightarrow P$ splits $\forall A$ iff $P$ is projective. It's also not hard to ensure that the assumptions of the proposition hold.
That subobjects of $1$ generate follows immediately from the facts that any presheaf is a colimit of representables and sheafification $a$ preserves subobjects and colimits (in "Sheaves ..." it is at the end of III.6, p. 139): $P \simeq aiP \simeq a\varinjlim yU_i \simeq \varinjlim a yU_i, a yU_i \in Sub_{Sh}(1).$
To see that $Sub(E)$ is a complete Boolean algebra, note that, as in III.8 (21), it is indeed the case when a topology is the dense topology on a poset. This is exactly our case. Alternatively, you can for a subsheaf $Q \in Sub(E)$ construct its negation $\bar{Q}$ explicitly. The classifying map for $Q$, $S \to \Omega = \{ \text{principal sieves} \} ,$ assigns to $s \in S(A)$ the maximal algebra object $U \subset A$ on which the restriction of $s$ falls into $Q(-).$ (Note that $\Omega$ is a Boolean algebra.) Compose the classifying map with the negation map $\neg \colon \Omega\to \Omega$ to get the classifying map of $\bar{Q}$. It's not hard to see that $Q \lor \bar{Q} = E$.
Note: this answer assumes that $Sh(B)$ is the category of sheaves where $B$ is considered as a locale, i.e. the presheaves $\mathscr{F}$ on the poset category of $B$ such that whenever $\bigvee_{i\in I} c_i = b$ and $f_i \in \mathscr{F}(c_i)$ satisfy $f_i |_{c_i \wedge c_j} = f_j |_{c_i \wedge c_j}$, then there exists a unique $f \in \mathscr{F}(b)$ such that $f |_{c_i} = f_i$. If that does not match the definition the problem is using, let me know and I can delete the answer.
Here is an outline of the proof: suppose we have an epimorphism $\phi : \mathscr{G} \to \mathscr{G}''$ and a morphism $g : \mathscr{F} \to \mathscr{G}''$. We now consider the partial order where objects are ordered pairs of a subsheaf $\mathscr{F}' \subseteq \mathscr{F}$ and a lifting $\tilde g : \mathscr{F}' \to \mathscr{G}$ such that $\phi \circ \tilde g = g |_{\mathscr{F}'}$; and the ordering is given by extension i.e. $(\mathscr{F}_1', \tilde g_1) \le (\mathscr{F}_2', \tilde g_2)$ is and only if $\mathscr{F}_1' \subseteq \mathscr{F}_2'$ and $\tilde g_1 = \tilde g_2 |_{\mathscr{F}_1'}$.
First, it is relatively straightforward to show that any chain $C$ in this partial order has an upper bound where $\mathscr{F}' = \bigcup_{(\mathscr{F}_i', \tilde g_i) \in C} \mathscr{F}_i'$ and $\tilde g$ is given by gluing the $\tilde g_i$. Therefore, by Zorn's Lemma, the partial order has a maximal element. We claim that this maximal element has $\mathscr{F}' = \mathscr{F}$, so that the $\tilde g$ component will be the desired lifting $\mathscr{F} \to \mathscr{G}$.
To see this, suppose to the contrary that for some $b \in B$ and some $x \in \mathscr{F}(b)$, $x \not\in \mathscr{F}'(b)$. Then let $c \le b$ be the maximum element such that $x |_c \in \mathscr{F}'(c)$, and let $c' := b \wedge \overline{c}$. Then since $\mathscr{G} \to \mathscr{G}''$ is an epimorphism, there exists a covering of $d_i \le c'$ with $\bigvee_{i\in I} d_i = c'$ and sections $y_i \in \mathscr{G}(d_i)$ such that $\phi(y_i) = f(x)|_{d_i}$. Since $c' \ne \bot$, we must have some $d_i \ne \bot$. Then, for this $i$, we can form an extension of $\tilde g$ to $\mathscr{F}' \cup \{ x|_{d_i} \}$ with $\tilde g'(x|_{d_i}) = y_i$ (where the essential point is that the union is a disjoint union since $d_i \wedge c = \bot$), which contradicts the assumed maximality of $(\mathscr{F}', \tilde g)$.