Proving NP-completeness of a surveilled graph problem
Let $\langle \mathcal{S}, \mathcal{I} \rangle$ be an instance of exact cover. Here $\mathcal{I}$ is a set of items and $\mathcal{S} \subseteq 2^\mathcal{I}$ is a collection of subsets of $\mathcal{I}$.

The goal is to decide whether there is a subset $S$ of $\mathcal{S}$ such that every item in $\mathcal{I}$ belongs to exactly one set in $S$. This is a well-known NP-complete problem.

To reduce exact cover to your problem create a graph $G = (V, E, A, g)$ as follows:


$V = \mathcal{S} \cup \mathcal{I}$.
$E = \{ (u,v) \in \mathcal{S} \times \mathcal{I} \mid v \in u \}$.
$A = \mathcal{I}$
$\forall u \in a, g(u)=1$.


If $S$ is an exact cover for $\langle \mathcal{S}, \mathcal{I} \rangle$, then $B=S$ is a solution for your original problem since $S \cap \mathcal{I} = B \cap A = \emptyset$ and each item in $\mathcal{I}=A$ has exactly one neighbor in $S = B$.

On the other hand, a solution $B$ to your original problem is such that $B \cap A = B \cap \mathcal{I} = \emptyset$. Hence $B \subseteq \mathcal{S}$ and eacn item in $A = \mathcal{I}$ has exactly one neighbor in $B$. Therefore $B$ is an exact cover for $\langle \mathcal{S}, \mathcal{I} \rangle$.