Explanation

(d)
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In $p$-dichlorobenzene, two $C - Cl$ dipole cancel each other
$\therefore\mu = 0$
In,$\ o$-dichlorobenzene, two $C - C$l dipoles (say $x$) are inclined at an angle of $60{^\circ}$.Therefore, according to parallelogram law of forces, the resultant
$= \sqrt{x^{2} + x^{2} + 2x \times \cos{60{^\circ}}}$
$= \sqrt{x^{2} + x^{2} + {2x}^{2} \times 1/2}$
$= \sqrt{3x^{2}} = \sqrt{3x}$
In $m$-dichlorobenzene, the two dipoles are inclined to each other at an angle of $120{^\circ}$, therefore, resultant
$= \sqrt{x^{2} + x^{2} + 2x \times \cos{120{^\circ}}}$
$= x^{2} + x^{2} + {2x}^{2} \times ( - 1/2)$
$= \sqrt{x^{2}} = x$
Thus ,the decreasing order of dipole moments: $o > m > p$