Let $X$ be a set.
Map I
We define a map
\[ \Phi _{X,Y}\colon \mathsf{SupLat}((\mathcal{P}(X),\subset ),(Y,\preceq )) \to \mathsf{Sets}(X,Y) \]
as in the statement, i.e. by
\[ \Phi _{X,Y}(f)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f\circ \chi _{X} \]
for each $f\in \mathsf{SupLat}((\mathcal{P}(X),\subset ),(Y,\preceq ))$.
Map II
We define a map
\[ \Psi _{X,Y}\colon \mathsf{Sets}(X,Y)\to \mathsf{SupLat}((\mathcal{P}(X),\subset ),(Y,\preceq )) \]
as in the statement, i.e. by
for each $f\in \mathsf{Sets}(X,Y)$.
Invertibility I
We claim that
\[ \Psi _{X,Y}\circ \Phi _{X,Y}=\operatorname {\mathrm{id}}_{\mathsf{SupLat}((\mathcal{P}(X),\subset ),(Y,\preceq ))}. \]
We have
\begin{align*} [\Psi _{X,Y}\circ \Phi _{X,Y}](f) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\Psi _{X,Y}(\Phi _{X,Y}(f))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\Psi _{X,Y}(f\circ \chi _{X})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{Lan}}_{\chi _{X}}(f\circ \chi _{X})\\ \end{align*}
for each $f\in \mathsf{SupLat}((\mathcal{P}(X),\subset ),(Y,\preceq ))$. We now claim that
\[ \operatorname {\mathrm{Lan}}_{\chi _{X}}(f\circ \chi _{X})=f \]
for each $f\in \mathsf{SupLat}((\mathcal{P}(X),\subset ),(Y,\preceq ))$. Indeed, we have
\begin{align*} \left[\operatorname {\mathrm{Lan}}_{\chi _{X}}(f\circ \chi _{X})\right](U) & = \bigvee _{x\in U} f(\chi _{X}(x))\\ & = f\left(\bigvee _{x\in U}\chi _{X}(x)\right)\\ & = f\left(\bigcup _{x\in U}\left\{ x\right\} \right)\\ & = f(U)\end{align*}
for each $U\in \mathcal{P}(X)$, where we have used that $f$ is a morphism of suplattices and hence preserves joins for the second equality. This proves our claim. Since we have shown that
\[ [\Psi _{X,Y}\circ \Phi _{X,Y}](f)=f \]
for each $f\in \mathsf{SupLat}((\mathcal{P}(X),\subset ),(Y,\preceq ))$, it follows that $\Psi _{X,Y}\circ \Phi _{X,Y}$ must be equal to the identity map $\operatorname {\mathrm{id}}_{\mathsf{SupLat}((\mathcal{P}(X),\subset ),(Y,\preceq ))}$ of $\mathsf{SupLat}((\mathcal{P}(X),\subset ),(Y,\preceq ))$.
Invertibility II
We claim that
\[ \Phi _{X,Y}\circ \Psi _{X,Y}=\operatorname {\mathrm{id}}_{\mathsf{Sets}(X,Y)}. \]
We have
\begin{align*} [\Phi _{X,Y}\circ \Psi _{X,Y}](f) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\Phi _{X,Y}(\Psi _{X,Y}(f))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\Phi _{X,Y}(\operatorname {\mathrm{Lan}}_{\chi _{X}}(f))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{Lan}}_{\chi _{X}}(f)\circ \chi _{X}\end{align*}
for each $f\in \mathsf{Sets}(X,Y)$. We now claim that
\[ \operatorname {\mathrm{Lan}}_{\chi _{X}}(f)\circ \chi _{X}=f \]
for each $f\in \mathsf{Sets}(X,Y)$. Indeed, we have
\begin{align*} [\operatorname {\mathrm{Lan}}_{\chi _{X}}(f)\circ \chi _{X}](x) & = \bigvee _{y\in \left\{ x\right\} }f(y)\\ & = f(x)\end{align*}
for each $x\in X$. This proves our claim. Since we have shown that
\[ [\Phi _{X,Y}\circ \Psi _{X,Y}](f)=f \]
for each $f\in \mathsf{Sets}(X,Y)$, it follows that $\Phi _{X,Y}\circ \Psi _{X,Y}$ must be equal to the identity map $\operatorname {\mathrm{id}}_{\mathsf{Sets}(X,Y)}$ of $\mathsf{Sets}(X,Y)$.
Naturality for $\Phi $, Part I
We need to show that, given a function $f\colon X\to X'$, the diagram commutes. Indeed, we have
\begin{align*} [\Phi _{X,Y}\circ \mathcal{P}_{!}(f)^{*}](\xi ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\Phi _{X,Y}(\mathcal{P}_{!}(f)^{*}(\xi ))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\Phi _{X,Y}(\xi \circ f_{!})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(\xi \circ f_{!})\circ \chi _{X}\\ & = \xi \circ (f_{!}\circ \chi _{X})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle (\dagger )}}{=}}}\xi \circ (\chi _{X'}\circ f)\\ & = (\xi \circ \chi _{X'})\circ f\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\Phi _{X',Y}(\xi )\circ f\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f^{*}(\Phi _{X',Y}(\xi ))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[f^{*}\circ \Phi _{X',Y}](\xi ), \end{align*}
for each $\xi \in \mathsf{SupLat}((\mathcal{P}(X'),\subset ),(Y,\preceq ))$, where we have used Item 1 of Proposition 4.5.4.1.3 for the fifth equality above.
Naturality for $\Phi $, Part II
We need to show that, given a morphism of suplattices
\[ g\colon (Y,\preceq _{Y})\to (Y',\preceq _{Y'}), \]
the diagram
commutes. Indeed, we have
\begin{align*} [\Phi _{X,Y'}\circ g_{!}](\xi ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\Phi _{X,Y'}(g_{!}(\xi ))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\Phi _{X,Y'}(g\circ \xi )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(g\circ \xi )\circ \chi _{X}\\ & = g\circ (\xi \circ \chi _{X})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}g\circ (\Phi _{X,Y}(\xi ))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}g_{!}(\Phi _{X,Y}(\xi ))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[g_{!}\circ \Phi _{X,Y}](\xi ). \end{align*}
for each $\xi \in \mathsf{SupLat}((\mathcal{P}(X),\subset ),(Y,\preceq ))$.
Naturality for $\Psi $
Since $\Phi $ is natural in each argument and $\Phi $ is a componentwise inverse to $\Psi $ in each argument, it follows from Chapter 11: Categories, Item 2 of Proposition 11.9.7.1.2 that $\Psi $ is also natural in each argument.
