Answer
a. Yes.
b. 20.4 MeV.
Work Step by Step
a. Check the sign of the Q-value, using data from Appendix B. We use atomic masses.
$$Q=\left( m_p+m(^{7}_{3}Li)-m(^{4}_{2}He)-m_{\alpha} \right)c^2$$
$$ =\left(1.007825u+7.016003u-2(4.002603)u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$=17.346MeV$$
This is positive, so it can occur without any energy input.
b. The total KE of the products is the Q-value plus the KE brought in by the incoming proton: 17.346MeV + 3.1 MeV = 20.4 MeV.
