Dieckmann condensation's final protonation step
why does the enol tautomerizes to the keto form?


In 1,3-dicarbonyl systems, like the one pictured above, we would generally expect to have a measurable amount of the enol form present.  The keto-enol interconversion is an equilibrium and ratio of keto and enol will be guided by thermodynamic factors.  Let's take a look.

If we compare the structural elements in the keto and enol forms, we see that the differences between the two tautomers are as follows:


The keto form has an extra carbonyl (178), a $\ce{C-C}$ single bond (83) and a $\ce{C-H}$ bond (99)
The enol form has a $\ce{C-O}$ bond (86), a $\ce{O-H}$ bond (111) and a $\ce{C=C}$ double bond (146)


The numbers in parenthesis are standard bond energies in kcal/mol.  If we add up the keto values we get 360 kcal/mol; while for the enol it totals to 343 kcal/mol.  The numbers strongly favor the keto form (by 17 kcal/mol).  That is why is most simple systems the keto form strongly predominates.

However as you point out, both resonance and hydrogen bonding can further stabilize the enol form in the current example.

The hydrogen bond strength in $\ce{O..H-O}$ systems is typically 5-6 kcal/mol. In this system, with an intramolecular bond and a 6-membered transition state the hydrogen bond strength is likely closer to 9-10 kcal/mol. Resonance stabilization in 1,3-pentadiene is approximately 6.2 kcal/mol. Due to increased polarization in the enol from the carbonyl replacing a carbon-carbon double bond, the resonance stabilization probably increases to something around 8 kcal/mol.

A 9 kcal/mol effect due to hydrogen bonding and a 8 kcal/mol push due to resonance, shifts the 17 kcal/mol stabilization of the keto form to something very close to where the keto and enol forms have comparable energy. Therefore, in  1,3-dicarbonyl compounds we would expect a delicate balance between keto and enol forms, where both forms may be detectable.