Answer
$\dfrac{\sqrt2}{4}$
Work Step by Step
Recall the rule: $(p \circ g)(x)=p \ [(g(x)] $
Let us consider: $p(x)=\frac{x}{2}$ and $g(x)=\cos x$
Thus, we have:
$(p\circ g)(315^\circ)=p [g(315^\circ)] \\ =p(\cos{315^\circ}) \\ =p (\dfrac{\sqrt2}{2}) \\ \\ =\dfrac{(\frac{\sqrt 2}{2}) }{2} \\ =\dfrac{\sqrt2}{4}$