Explanation

(b) First we have to find a point where the resultant field due to both is zero. Let the point $P$ be at a distance $x$ from centre of bigger star.
$\Rightarrow \frac{G(16M)}{x^{2}} = \frac{GM}{(10a - x)^{2}} \Rightarrow x = 8a\ $(from $O_{1}$)
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$ie,$ once the body reaches $P$, the gravitational pull of attraction due to $M$ takes the lead to make $m$ move towards it automatically as the gravitational pull of attraction due to 16 $M$vanishes $ie$, a minimum KE or velocity has to be imparted to $m$ from surface of $16M$ such that it is just able to overcome the gravitational pull of $16M$. By law of conservation of energy.
(Total mechanical energy at $A$) = (Total mechanical energy at $P$)
$\Rightarrow \frac{1}{2}mv_{\min}^{2} + \left\lbrack \frac{G(16M)m}{2a} - \frac{GMm}{8a} \right\rbrack$
$= 0 + \left\lbrack \frac{GMm}{2a} - \frac{G(16M)m}{8a} \right\rbrack$
$\Rightarrow \frac{1}{2}mv_{\min}^{2} = \frac{GMm}{8a}(45) \Rightarrow v_{\min} = \frac{3}{2}\sqrt{\frac{5GM}{a}}$