Here's an answer, we can just verify that $(-)\int \newcommand\A{\mathcal{A}}\A$ is left adjoint to $\newcommand\set[1]{\left\{{#1}\right\}}\set{\A,(-)}$ by direct computation.
Side note, the definition of $\mathcal{D}\int A$ as given doesn't make sense for general $\mathcal{D}$, we need to be keeping track of which object of $\mathcal{D}$ over $n\in \mathbb{P}$ we're at, so objects should be pairs $(X,[A_i])$ of an object over $n$ and a string of $n$ objects of $\A$.
Suppose $F: \newcommand\D{\mathcal{D}}\D\int \A\to \newcommand\B{\mathcal{B}}\B$ is a functor. By definition, this consists of the following data, for each pair $(D,[A_i])$ of an object of $\D$ and a string of length $n$, an object $F(D,[A_i])$ in $\B$, and for each morphism $(\alpha, f_1,\ldots,f_n) : (D,[A_i])\to (D',[B_i])$, an appropriate morphism $F(\alpha,f_1,\ldots,f_n)$ such that the composition law is satisfied:
$$
F(\beta,g_1,\ldots,g_n) \circ F(\alpha,f_1,\ldots, f_n) = F(\beta\alpha,g_{\alpha(1)}f_1,\ldots,g_{\alpha(n)}f_n).
$$
Side note: This looks a lot like a categorified wreath product to me.
In particular, this is a sort of twisted product, so we should expect it to be left adjoint to some sort of twisted hom, which is what $\set{\A,\B}$ should be.
Let $\pi : \D\to \mathbb{P}$ be the structure map for $\D$.
On the other hand, if $G : \D\to \set{\A,\B}$ is a functor of categories over $\mathbb{P}$, then this consists of for each $D\in \D$, a choice of an object $GD = (n,G_D : \A^n\to \B)$, where we must have $n = \pi D$, and for each $\alpha :D\to D'$, a morphism $(\sigma, \eta_\alpha : \sigma \cdot G_D\to G_{D'})$, where we are forced to have
$\sigma = \pi(\alpha)$.
Now the functors $G_D$ for each $D$ themselves consist of the following data, for each string $[A_i]$ of $n$ objects of $\A$, an object $G_D([A_i])$ of $\B$, and for each morphism $(f_1,\ldots,f_n) : [A_i]\to [B_i]$ a morphism $G_D(f_1,\ldots,f_n) : G_D([A_i])\to G_D([B_i])$, subject to the composition rule.
The $\eta_\alpha$s are a family of morphisms $G_D([A_{\sigma(i)}])\to G_{D'}([A_i])$
such that for all $(f_1,\ldots,f_n)$ the following diagram commutes
$$
\require{AMScd}
\begin{CD}
G_D([A_{\sigma(i)}]) @>G_D(f_{\sigma(1)},\ldots,f_{\sigma(n)})>> G_D([B_{\sigma(i)}])
\\
@V\eta_{\alpha}VV @VV\eta_{\alpha}V
\\
G_{D'}([A_i]) @>G_{D'}(f_1,\ldots,f_n)>> G_{D'}([B_i])
\\
\end{CD}
$$
Of course the diagram still commutes if we permute the maps by $\sigma^{-1}$, and we get
$$
\require{AMScd}
\begin{CD}
G_{D}([A_i]) @>G_{D}(f_1,\ldots,f_n)>> G_{D}([B_i])
\\
@V\eta_{\alpha}VV @VV\eta_{\alpha}V
\\
G_{D'}([A_{\sigma^{-1}(i)}]) @>G_{D'}(f_{\sigma^{-1}(1)},\ldots,f_{\sigma^{-1}(n)})>> G_{D'}([B_{\sigma^{-1}(i)}])
\\
\end{CD}
$$
However, there's actually an indexing discrepancy between the indices of the $B_i$ here and the $B_i$ in the definition of $\D\int \A$. Namely, the maps from $[A_i]$ to $[B_i]$ in $\D\int\A$ are $f_i: A_i\to B_{\sigma(i)}$.
Thus we see that if we reindex accordingly, for a fixed choice of $(\alpha,f_1,\ldots,f_n)$, with $f_i:A_i\to B_{\sigma(i)}$, the diagonal map is a map $G_D([A_i])\to G_{D'}([B_i])$.
In other words, we have recovered the data of a functor $F:\D\int\A\to \B$, for each object $(D,[A_i])$ we define $F(D,[A_i])=G_D([A_i])$, and for each $(\alpha,f_1,\ldots,f_n)$, we define $F(\alpha,f_1,\ldots,f_n)$ to be the diagonal map constructed above.
Conversely, if we start with such a functor $F$, we can go backwards and produce the data of a functor $G$. We define $G_D$ to be $F(D,-)$, with $G_D$ defined on morphisms by $G_D(f_1,\ldots,f_n) = F(1_D,f_1,\ldots,f_n)$. Then the natural transformations $\eta_\alpha$ should be $F(\alpha,1_{A_{\sigma^{-1}(1)}},\ldots,1_{A_{\sigma^{-1}(n)}})$. (I think, not quite sure if the $\sigma^{-1}$ is correct, but it's getting a bit late, so I'll just let you check that detail.)
This establishes a bijection of collections of functors. It shouldn't be too bad to show that it is a natural bijection.
End Note I feel like there ought to be a general categorical notion underlying both constructions, and if someone knows what that is, I'd love to hear about it.
Best Answer
This looks right.
Essentially what you're saying is that every $f=(f_1,f_2) : (A_1,A_2) \to (B_1,B_2)$ in $\mathbf{Set} \times \mathbf{Set}$ induces a function $f_1 \times f_2 : A_1 \times A_2 \to B_1 \times B_2$ in $\mathbf{Set}$, where $f_1 \times f_2 = \langle f_1 \circ \pi_{A_1}, f_2 \circ \pi_{A_2} \rangle$, but not every function $A_1 \times A_2 \to B_1 \times B_2$ is of this form.
Or even more concisely, the product functor $\mathbf{Set} \times \mathbf{Set} \to \mathbf{Set}$ is not full.