Equilibrium: JEE Advanced 2021 Chemistry question

Correct match of the $\mathbf{C}-\mathbf{H}$ bonds (shown in bold) in Column $\mathbf{J}$ with their BDE in Column $\mathbf{K}$ is \begin{center} \begin{tabular}{|c|c|} \hline $\begin{array}{l}\text { Column J } \\ \text { Molecule }\end{array}$ & $\begin{array}{c}\text { Column K } \\ \text { BDE }\left(\text { kcal mol }^{-1}\right)\end{array}$ \\ \hline (P) $\mathbf{H}-\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$ & (i) 132 \\ \hline (Q) $\mathbf{H}-\mathrm{CH}_{2} \mathrm{Ph}$ & (ii) 110 \\ \hline (R) $\mathbf{H}-\mathrm{CH}=\mathrm{CH}_{2}$ & (iii) 95 \\ \hline (S) $\mathrm{H}-\mathrm{C} \equiv \mathrm{CH}$ & (iv) 88 \\ \hline \end{tabular} \end{center}