Explicitely, $B_\kappa$ can be taken to be any (non-trivial) Boolean locale that collaps"collaps the cardinal $\kappa$ to $\omega$" I mean by that, if $p:B_\kappa \to 1$ denotes the unique map, then $p^*\kappa \simeq p^* \omega$ as sheaves over $B_\kappa$ . For example, $B_\kappa$ iscan be taken to be the double negation sublocale of the locale of injective functions $\omega \to \kappa$. The fact that $B_\kappa$ is non trivial then follows from the fact that it is dense in thethis locale of injective function. (See details in edit below)
And a locale $X$ having functions to all the $B_\kappa$ would collapse all infinite cardinals at the same time (in the sense that all $p^*\kappa$ for infinite cardinal $\kappa$ would be isomorphic), which is impossible as a locale can't collapse to $\omega$ cardinal much larger than itself. THe following are very loose bound that shows it, though experts on forcing surely known much better bounds:
If (I don't have$X$ is a non-degenerate locales, then the precise obstruction/bound in mindtotal number of locale section of $p^* \kappa$ is larger than $\kappa$ as every element of $\kappa$ gives a globale section, and it is smaller than the function space $\kappa^{\mathcal{O}(X) \times 2^{\mathcal{O}(X)}}$ as every locale section can be written as: you chose a cover of its domain of definition (a special subset of $\mathcal{O}(X)$) and then for thiseach element of that subset, if some one know themyou chose an element of $\kappa$.
So for any locale $X$, feel freepicking a $\kappa$ to edit or comment with some details)be larger than $\omega^{\mathcal{O}(X) \times 2^{\mathcal{O}(X)}}$ we get that $p^* \omega$ can't have as much section as $p^* \kappa$ and hence they can't be isomorphic, so $X$ can't collaps any cardinal bigger than this to $\omega$.