1. Idea used: acidity of a proton depends on the stability of the conjugate base formed after removing that proton.

2. For the marked protons $H_A, H_B, H_C, H_D$, we compare how well the negative charge is stabilized after deprotonation.

3. Most acidic proton: $H_C$
   $H_C$ is the proton whose removal gives an enolate/conjugate base stabilized most strongly by resonance with the carbonyl group and by the strongest electron-withdrawing environment. Hence it is the most acidic.

4. Next: $H_D$
   Removal of $H_D$ also gives a resonance-stabilized conjugate base, but the stabilization is less than for $H_C$. Therefore, $H_C > H_D$

5. Compare $H_A$ and $H_B$
   Between these two, the conjugate base from $H_A$ is more stabilized than that from $H_B$ (due to better inductive/resonance support), so $H_A > H_B$

6. Combining all comparisons, $H_C > H_D > H_A > H_B$

7. Option matching: this corresponds to Option D.

Final answer: $\boxed{H_C > H_D > H_A > H_B}$