$a=\large\frac{dv}{dt}$
Force due to power P is
$F=\large\frac{p}{v}$$;f=\mu R$
$F-mg \sin \theta-\mu R=ma$
$\large\frac{P}{v}$$-mg \sin \theta - \mu mg \cos \theta=m \large\frac{dv}{dt}$
Velocity is maximum when $\large\frac{dv}{dt}$$=0$
$\large\frac{P}{v_{\Large max}}$$= mg \sin \theta +\mu mg \cos \theta$
$v_{max}= \large\frac{P}{mg \sin \theta+\mu mg \cos \theta}$
Hence a is the correct answer