I think we can show that inequality holds for any commutative quantale, and not just for \$$\mathbf{Bool}\$$.
We start from the definition of enriched functor, since the mapping \$$\Phi\$$ is a \$$\mathcal{V}\$$-functor between the \$$\mathcal{V}\$$-categories \$$\mathcal{X}^{\text{op}} \times \mathcal{Y}\$$ and \$$\mathcal{V}\$$:

$(\mathcal{X}^{\text{op}} \times \mathcal{Y})((x, y), (x', y')) \le \mathcal{V}(\Phi(x, y), \Phi(x', y')).$

We use the definition of the product of \$$\mathcal{V}\$$-categories to get

$\mathcal{X}^{\text{op}}(x, x') \otimes \mathcal{Y}(y, y') \le \mathcal{V}(\Phi(x, y), \Phi(x', y'))$

and the definition of the opposite of \$$\mathcal{V}\$$-category to get

$\mathcal{X}(x', x) \otimes \mathcal{Y}(y, y') \le \mathcal{V}(\Phi(x, y), \Phi(x', y')).$

By remark 2.86 in the book, we know that we can enrich \$$\mathcal{V}\$$ in itself using the closed property

$\mathcal{V}(v, w) = v \multimap w$

which gives

$\mathcal{X}(x', x) \otimes \mathcal{Y}(y, y') \le \Phi(x, y) \multimap \Phi(x', y')$

By applying the definition of the closed element (or hom-object)

$\mathcal{X}(x', x) \otimes \mathcal{Y}(y, y') \otimes \Phi(x, y) \le \Phi(x', y')$

and using commutativity of the quantale \$$\mathcal{V}\$$, we finally show that

$\mathcal{X}(x', x) \otimes \Phi(x, y) \otimes \mathcal{Y}(y, y') \le \Phi(x', y').$