The sum of two forces → P and → Q is → R such that | → R | = | → P...

So angle between $\left(2\overset{\to}{P} + \overset{\to}{Q}\right)$ and $\overset{\to}{Q}$ is ${90}^{o}$

Alternate solution

\left|\overset{\to}{\text{P}} + \overset{\to}{\text{Q}}\right| = \left|\overset{\to}{\text{P}}\right|
${P}^{2} + {Q}^{2} + 2PQcos \theta = {P}^{2}$
$⇒ Q + 2PQcos \theta = O$
$⇒ cos \theta = -$\dfrac{Q}{2P}

$tan \alpha =$\dfrac{2Psin \theta}{2Pcos \theta + Q} =$\infty$
$∵ \left[2Pcos \theta + Q = 0\right]$
$\alpha = {90}^{o}$