Suppose $\mathcal{A}$ is the empty collection. I claim that $$B - \bigcap\limits_{A \in \mathcal{A}} A = \bigcup\limits_{A \in \mathcal{A}} (B - A).$$ Denote the set-theoretic universe by $\mathcal{U}$. By convention, $$\bigcap\limits_{A \in \mathcal{A}} A = \mathcal{U}.$$ So for $x$ to live in $B - \bigcap\limits_{A \in \mathcal{A}}$ is to say that $x$ lives in $B - \mathcal{U}$. To live in $B$ requires living in $\mathcal{U}$, so the LHS is the empty set.
On the right-hand side, we have $$x \in \bigcup\limits_{A \in \mathcal{A}} (B - A) \iff \exists A \in \mathcal{A}, \; x \in (B - A).$$ But since there do not exist $A \in \mathcal{A}$, the right-hand side is false for any $x \in \mathcal{U}$, so the RHS is also the empty set.
Is this correct?