First look the bond energies. Breaking of the bonds creates any unstable compound.
For example in the above question, Hb-C bond breaks.
It results to the formation of a carbocation on the double bond, which makes the compound unstable.  So, C-Hb has highest bond energy.
Then check if breaking of the bond forms any stable carbocation.
They will easily break the bond to attain that stable structure.
So, their bond energy will be less. For ex in the above compound Ha-C, Hc-C, Hd-C on breaking form stable carbocation.
Between Ha and Hc, Ha involves in conjugation where as Hc doesn't. Thus C-Ha bond energy is less than that of C-Hc bond energy.
Hence The decreasing order is Hb>Hc>Ha>Hd.

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Order of bond energy of protons Ha,Hb,HC,Hd

The bond energy of a proton is determined by the strength of the bond it forms with another atom. This strength is determined by the electronegativity of the atom to which the proton is bonded.

Ha > Hb > HC > Hd

Explanation:

1. Electronegativity of the atoms:

The electronegativity of the atoms to which the protons are bonded plays a significant role in determining the bond energy. The greater the electronegativity of the atom, the stronger the bond.

2. Ha:

Ha is bonded to the most electronegative atom among the four options. This means that the bond energy of Ha will be the highest among the four.

3. Hb:

Hb is bonded to an atom that is less electronegative than the one to which Ha is bonded. This means that the bond energy of Hb will be lower than that of Ha.

4. HC:

HC is bonded to an atom that is less electronegative than the one to which Hb is bonded. Therefore, the bond energy of HC will be lower than that of Hb.

5. Hd:

Hd is bonded to the least electronegative atom among the four options. This means that the bond energy of Hd will be the lowest among the four.

In conclusion, the decreasing order of bond energy of protons Ha, Hb, HC, Hd is Ha > Hb > HC > Hd.