Answer
$P=256N$
Work Step by Step
We have $\sum W=\sum F\times s$, so $\sum W=0$ when $\sum F=0$, or the forces cancel each other out.
There are only 2 horizontal forces acting in the opposite direction and influencing the horizontal movement of the crate: $P\cos30^o$ and $f_k$. Therefore $$\sum F=P\cos30^o-f_k=0.866P-f_k=0 (1)$$
We now take a look at $f_k$ $$f_k=\mu_kF_N=0.2F_N$$
There are 3 vertical forces in play here: $F_N$ pointing upward while $mg$ and $P\sin30$ point downward. As there is no vertical acceleration, the upward forces balance the downward ones: $$F_N=mg+P\sin30=100\times9.8+P\sin30=980+0.5P$$
Therefore, $$f_k=0.2(980+0.5P)=196+0.1P$$
Plug this into (1): $$0.866P-196-0.1P=0$$ $$0.766P=196$$ $$P=256N$$
