$P(getting\ a\ head\ from \ coin \ A) = \frac{1}{4}$
$P(getting\ a\ head\ from \ coin \ B) = \frac{3}{4}$
Probability of choosing $A =P(A)= \frac{1}{2}$
Probability of choosing $B =P(B) = \frac{1}{2}$
If we choose $A$ the probability of getting head twice $=P(H/A)= \frac{1}{4} * \frac{1}{4} = \frac{1}{16}$
If we choose $B$ the probability of getting head twice $=P(H/B)= \frac{3}{4} * \frac{3}{4} = \frac{9}{16}$
$P(B/H) = ? $
So using Baye's theorem,
$P(B/H) = \frac{P(H/B) *P(B)}{ P(H/A) *P(A) + P(H/B) *P(B) } = \frac{\frac{9}{16} *\frac{1}{2}}{ \frac{1}{16} *\frac{1}{2} + \frac{9}{16} *\frac{1}{2} } = \frac{9}{10}$
So option $D$ is the correct answer.