Answer
d) is correct.
Work Step by Step
The work done by a force $F$ on the box is $$W=(F\cos\theta)s$$
Since $P$ propels the box's motion in the same direction as $\vec{v}$, $\cos\theta=\cos0=1$, therefore, $W_P=Ps\gt0$
Kinetic friction $f_k$ always opposes the motion in its opposite direction, so $\cos\theta=\cos180=-1$, meaning $W_f=-f_ks\lt0$
$F_N$ and $mg$ are vertical forces, while the motion is horizontal, so $\cos\theta=\cos90=0$ and $W_N=W_{mg}=0$
So d) is correct.
