I did the assignment for both tautomers, ketone 1 and enol 1a. You can determine the ratio of the KET of dimedone (1) in CDCl3 (as an example) just by integration from a simple 1D experiment. For this, I divided the integrals of the (CH3)2CCH2 groups of 1 by the integral of the same protons in 1a, namely the singlets at 2.85 and 2.59 ppm. The result gives a keto:enol ratio of 2.9:1 (4.0/1.4 = 2.9). This corresponds to a value of 74% in favor of the keto form, which is slightly higher but in good agreement with what we should expect.[2]
The signals of the enol species 1a in the 1H NMR spectrum correspond to a symmetric molecule. The reason for this is that a rapid intramolecular proton exchange from one oxygen atom to the other oxygen center is taking place. As this dynamic process is faster than the NMR time scale, both species are in equilibrium and the two CH2 groups of the enol 1a are observed as a broad singlet.[2]
By the way, if D2O was used as solvent instead of chloroform, the signal of the acidic proton would disappear over time, as H/D exchange would occur (for more details see James’ blog post To D2O or not to D2O?).
References
[1]http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf (accessed November 2018).
[2]J. Clayden, N. Greeves, S. Warren and P. Wothers, Organic Chemistry, Oxford University Press, Oxford/Berlin, 2001, p. 530.