(d)
We know that,
$\tan{(A - B) = \frac{\tan{A - \tan B}}{1 + \tan{A\tan B}}}$
$= \frac{\frac{x\sqrt{3}}{2k - x} - \frac{2x - k}{k\sqrt{3}}}{1 + \frac{x\sqrt{3}}{2k - x}.\frac{2x - k}{k\sqrt{3}}} = \frac{1}{\sqrt{3}}$
$\Rightarrow A - B = 30{^\circ}$