(I know the OP said to disregard this question, but I discovered some nice things that I'd like to share in this answer.)
Notations. Let me start by introducing a slightly different notation, writing $D^\#$ instead of $\mathcal{A}_D$: $ D^\# = \{ V,f : V \in D : f^{-1}[V] \} $ Read this as "the set which, for each $V$ and $f$ such that $V \in D$, contains $f^{-1}[V]$". Here, and in the rest of the post, we let $f$ range implicitly over $\{\varphi_i\}_{i \in \Lambda}$.
We use $D^\cup$ and $D^\cap$ for the set of all arbitrary unions and of finite intersections of elements of $D$, respectively: $ \begin{align} & D^\cup = \left\{ \mathscr{F} : \mathscr{F} \subseteq D : \bigcup \mathscr{F} \right\} \\ & D^\cap = \left\{ F,G : F,G \in D : F \cap G \right\} \\ \end{align} $ (See also another answer of mine.) This means that we can write "the topology generated by $B$" as $B^{\cap\cup}$.
Problem statement. Using these notations, we are asked whether, for all $B$ and $\tau$ with $\tau = B^{\cap\cup}$, we have $\tau^{\#\cap\cup} = B^{\#\cap\cup}$; or equivalently, whether for all $B$ $ (0) \;\;\; B^{\cap\cup\#\cap\cup} = B^{\#\cap\cup} $ Proof. The only thing we know about $^\#$, $^\cap$, and $^\cup$ are their definitions: these apply $f^{-1}[\cdot]$, $\cap$, and $\cup$ many times. And looking at the latter operators, we know that the order of $f^{-1}[\cdot]$ and $\cap$ doesn't matter, and the same for $f^{-1}[\cdot]$ and $\cup$. (I will not prove those distribution properties here.) Therefore it seems likely that for all $D$ $ (1) \;\;\; D^{\cup\#} = D^{\#\cup} \\ (2) \;\;\; D^{\cap\#} = D^{\#\cap} \\ $ If we would be able to prove these, then $(0)$ directly follows from a simple calculation: $ \begin{align} & B^{\cap\cup\#\cap\cup} \\ = & \;\;\;\;\;\text{"using $(1)$ with $D := B^\cap$"} \\ & B^{\cap\#\cup\cap\cup} \\ = & \;\;\;\;\;\text{"using $(2)$ with $D := B$"} \\ & B^{\#\cap\cup\cap\cup} \\ = & \;\;\;\;\;\text{"$^{\cap\cup}$ is idempotent"} \\ & B^{\#\cap\cup} \\ \end{align} $ (I won't prove here that $^{\cap\cup}$ is idempotent.)
It turns out we can prove $(1)$, as follows: $ \begin{align} & D^{\cup\#} \\ = & \;\;\;\;\;\text{"definition of $^\#$"} \\ & \{ V,f : V \in D^\cup : f^{-1}[V] \} \\ = & \;\;\;\;\;\text{"write $V$ as a union using the definition of $^\cup$"} \\ & \{ \mathscr{F},f : \mathscr{F} \subseteq D : f^{-1}\left[\bigcup \mathscr{F} \right] \} \\ = & \;\;\;\;\;\text{"distribution property of $^{-1}$"} \\ & \{ \mathscr{F},f : \mathscr{F} \subseteq D : \bigcup \{ V : V \in \mathscr{F} : f^{-1}[V]\} \} \\ = & \;\;\;\;\;\text{"rewrite using the definition of $^{\cup}$"} \\ & \{ V,f : V \in D : f^{-1}[V] \}^{\cup} \\ = & \;\;\;\;\;\text{"definition of $^\#$"} \\ & D^{\#\cup} \\ \end{align} $ The proof for $(2)$ is very similar.
Observations. As a first observation, I quite like how the $^\#$ notation works out, and interacts with the $^\cap$ and $^\cup$ notations I'd invented earlier: $(0)$ brings out the structure of the problem in a way that is much clearer than writing $(\mathcal{A}_{B^{\cap\cup}})^{\cap\cup} = (\mathcal{A}_B)^{\cap\cup}$: after making the latter explicit, I suddenly realized that $\mathcal{A}$ is essentially a unary operator.
Also, I like how we've been able to avoid $\subseteq$, and instead have an equality proof. That is not what I expected when I started to work on this proof.
Finally, I dislike the set comprehension manipulation: the two steps above "using the definition of $^\cup$" are not really easy to follow. I don't yet see how to fix that.