Q value of the reaction is,

                                 Q = (2 × 4 × 7.06 – 7 × 5.6) MeV = 17.28 MeV



Applying conservation of energy for collision,

                                 K_p + Q = 2 K_α                 ....(i)

(Here, K_p and K_α are the kinetic energies of proton and α - particle respectively)

From the conservation of linear momentum (As there is no external force) ....(ii)

                               $[\mathrm{Here\; P}=\sqrt{2\mathrm{mk}}]$
                                ${\Rightarrow K}_p=16K_{\alpha }{\text{cos}}^2\theta =\left(16K_{\alpha }\right){\left(\frac{1}{4}\right)}^2\left(\text{as}{m}_{\alpha }=4m_p\right)$

∴                                 K_α =K_{p                 ....(iii)}

Solving eqs. (i) and (iii) with Q = 17.28 MeV


we get               K_p = 17.28 MeV