Inductive vs resonance effects and the acidity of phenol
The acidity of $\ce{A-H}$ is a measure of where the following equilibrium lies.

$$\ce{A-H &lt;=&gt; A^- + H+}$$

This means that we are comparing the relative stabilities of the products and reactants ($\ce{\Delta G=-RTlnK}$). 

In the present example (assuming your $\mathrm{p}K_\mathrm{a}$ for propen-2-ol is really being measured in water, note too that in water the relative concentration of this enol of acetone is ~ $10^{-8}$, how can you measure the $\mathrm{p}K_\mathrm{a}$ of something so dilute?) we are comparing the relative stability between [phenol and phenoxide] to the relative stability between [propen-2-ol and the corresponding anion].  Just because the two systems have the same difference in relative stabilities does not mean that phenol and propen-2-ol have the same stability.  In other words, just because the two systems have similar $\mathrm{p}K_\mathrm{a}$'s this cannot be interpreted to mean that the additional resonance structures in phenol and phenoxide are not significant contributors to both their stability and description.

EDIT: Jerepierre's comment as worked up by Dissenter (Addressing jerepierre's point)

in his edit Dissenter wrote, 


  Given this, we can conclude that the reactants are less stable in the acid-base sense than the products


I agree

Dissenter further posits,


  acetone enolate is less stable than phenoxide. Phenol is also less stable than acetone enol.


That's where I disagree.  We know that the reactants are less stable than the products, but I don't see how we know which reactant(s) is less stable than which of the products.

Let's say the reactants are $10\,\mathrm{kcal/mol}$ (made up number) less stable than the products.  Is phenol $5\,\mathrm{kcal/mol}$ less stable than phenoxide and acetone enolate $5\,\mathrm{kcal/mol}$ less stable than acetone enol?  Or is it $10$ and $0$ or $0$ and $10$, or $15$ and $-5$?  How can you break down the overall reactant/product difference of $10\,\mathrm{kcal/mol}$ further?