Answer by Gro-Tsen for Does $\neg(j^{**}(\bot))=\neg\neg(j^{*}(\bot))$ hold for all nuclei $j$ and nucleic/localic pseudocomplements $*$/$\neg$?

No, this does not hold in general. Here is a counterexample.

Let $L := \mathcal{O}(\mathbb{R})$ be the frame of open sets in $\mathbb{R}$. Let $D$ be a dense subset of $\mathbb{R}$ whose complement $E := \mathbb{R}\setminus D$ is also dense (e.g., $\mathbb{Q}$). Let $j$ be the nucleus associated with the subspace $D \subseteq \mathbb{R}$, that is, $j(V)$ (for $V\in L$) is the largest open set such that $U \cap D = V \cap D$, namely $\newcommand{\interior}{\operatorname{int}}\interior(V \cup (\mathbb{R}\setminus D)) = \interior(V\cup E)$. Symmetrically, let $k$ be the nucleus associated with the subspace $E \subseteq \mathbb{R}$, so $k(V) = \interior(V\cup D)$.

Claim:$k$ is the pseudocomplement $j^*$ of $j$ in the frame $N(L)$ of nuclei on $L$.

Proof. On the one hand, $j(V) \cap k(V) = \interior((V\cup D) \cap (V\cup E)) = V$, that is, $j\wedge k = \bot_{N(L)}$ in $N(L)$. This shows $k \leq j^*$ in $N(L)$. We now show the converse.

But by the formula recalled in the question,$$\begin{aligned}j^*(V) &= \bigwedge_{W\geq V} (j(W) \Rightarrow W)\\&= \interior\bigcap_{W\supseteq V}(j(W) \Rightarrow W)\\\end{aligned}$$where $W$ ranges over all open sets containing $V$, and $U\Rightarrow W$ denotes $\interior(W\cup(\mathbb{R}\setminus U))$ the Heyting operation on $L$.

Fix $V\in L$. If $x \not\in V\cup D$, consider $W := \mathbb{R}\setminus\{x\}$ (which is indeed open). We have $W \supseteq V$ as $x\not\in V$; and $j(W) = \interior(W\cup E) = \mathbb{R}$ (as $x\not\in D$), so that $(j(W) \Rightarrow W) = W$. So $x \not\in \bigcap_{W\supseteq V}(j(W) \Rightarrow W)$. This shows that $j^*(V)$ omits every $x \not\in V\cup D$, in other words, $j^*(V)$ is contained in $V\cup D$, and as it is open, it is contained in $\interior(V\cup D) = k(V)$. That is, $j^*(V) \subseteq k(V)$.

So we have shown both that $k \leq j^*$ and that $j^* \leq k$ in $N(L)$, as claimed. ∎

(This proof did not use the assumption that $D$ and its complement $E$ are dense, merely that $D$ is a subspace of a $\mathrm{T}_1$ topological space.)

Now since $D$ and $E$ play symmetric roles, $j^{**} = k^* = j$.

So in the titular question, $j^{**}(\bot_L) = j(\bot_L) = \interior(D) = \varnothing = \bot_L$ as $E$ is dense, and $j^*(\bot_L) = k(\bot_L) = \interior(E) = \varnothing = \bot_L$ as $D$ is dense. So the left hand side of $\neg(j^{**}(\bot_L))$ is $\top_L$ (that is, $\mathbb{R}$) and the right hand side is $\bot_L$, making the equality as false as it possibly can.