Angles

Since in a triangle, sum of all angle = 180

Therefore,

$\angle{P} + \angle{Q} = 360 - 2 \angle{BAD} - \angle{ABC} - \angle{ADC} = 180 + \angle{DCB} - ( \angle{BAD} + \angle{ADC} + \angle{ABC} )$ ...... (1)

Now,

$\angle{PRQ} = \angle{PCQ} - \frac{1}{2} ( \angle{P} + \angle{Q} )$

$\Rightarrow \angle{PRQ} = \angle{BCD} - \frac{1}{2} [ 180 + \angle{BCD} - ( \angle{BAD} + \angle{ABC} + \angle{ADC} )]$

$\Rightarrow \angle{PRQ} = \frac{1}{2} ( \angle{ABC} + \angle{ADC} + \angle{BAD} + \angle{BCD} ) - 90$

$\Rightarrow \angle{PRQ} = 180 - 90 = \large \boxed{90}$