Conserving energy, $(1/2)mv_a^2 - GMm/r_a = (1/2)mv_p^2 - GMm/r_p$.
Conserving moment of momentum, $r_av_a = r_pv_p$.

And since $GM = gR^2$,
$v_a^2 - 2gR^2/r_a = v_p^2 - 2gR^2/r_p$.

And since $v_ar_a = v_pr_p$,
$$v_a^2 - v_p^2 = 2gR^2(1/r_a - 1/r_p)\;,$$
and
$$v_a^2 - v_p^2 = 2gR^2(v_a/v_pr_p - v_p/v_pr_p)\;.$$

And applying the difference between 2 squares,
$$(v_a + v_p)(v_a - v_p) = 2gR^2(v_a/v_pr_p - v_p/v_pr_p)\;,$$
$$v_pr_p(v_a + v_p) = 2gR^2\;.$$

Applying the moment of momentum relationship again,
$$v_pr_p(v_a + v_ar_a/r_p) = 2gR^2\;,$$ so

$$v_av_p(r_a + r_p) = 2gR^2\;,$$ as required.

I found this simple formula that combines conservation of energy and moment of momentum while studying 1st year engineering, and decided to remember it in case a relevant question came up in the exam, rather than deriving the conservation from first principles, to save time!