Why is the pKa of a ketone higher than of water?
As I'm sure you know, (p)Ka is a measure of the equilibrium between an acid and its conjugate base (what you're calling NBE?). Both the conjugate bases of water (hydroxide) and a ketone (enolate) have a negative charge on the oxygen. I'm summarizing your question as: If both conjugate bases have a negative charge on the same atom type, but one base (the enolate) is stabilized by resonance, shouldn't that corresponding acid (ketone) be more acidic? (Correct me if I'm misreading you.)

I think the key here is what (p)Ka really represents. Fot water, with room temp pKa 15.7, the equilibrium favors water but some dissociation is present. For acetone, pKa 19.2, the equilibrium lies further on the non-dissociated side than for water. We take this to mean that acetone is a weaker acid that water. If we only looked at the stability of the conjugate bases, we might conclude the opposite (previous paragraph). But as pKa measures a balance between a conjugate pair, we must consider the acid forms as well. 

The ketone contains the very "stable" (energetically favorable) C=O carbonyl bond. When a ketone donates an alpha proton to become the enolate conjugate base, the carbonyl bond is disrupted. This is an extra energetic cost that is not present in the water equilibrium, and I propose it is the reason why the ketones tend to dissociate less than water. 

By this logic, water is a better acid than a ketone because acting as an acid requires a ketone to disrupt an energetically preferable C=O bond. Thus the balance lies closer to the undissociated side for a ketone (lower Ka, higher pKa) than for water (lower pKa), so we say that water is a stronger acid and a ketone is weaker. 

(One might argue that the carbonyl need not be disrupted, the lone pair could localize on the alpha carbon, but I hope we can agree such a carbanion is quite unstable, and a carbanion not preferable to breaking the carbonyl and localizing on the more electronegative oxygen.)