Answer
$\theta=43^o$
Work Step by Step
Kinetic friction $f_k$ has the following formula: $$f_k=\mu_kF_N$$
$F_N$ is a vertical force and hence the value of $F_N$ depends on the other vertical forces. Since there is no vertical motion whatsoever, these forces have to balance one another.
1) When the pulling force $P$ points horizontally, $P$ does not have a vertical component.
Therefore, there are only 2 vertical forces: $F_N$ and the box's weight $mg$. We would have $F_N=mg=150N$ and therefore, $f_1=150\mu_k$
2) When the pulling force $P$ points above the horizontal at an angle $\theta$, the vertical component $P_y=P\sin\theta$ and it points upward.
So vertically, there are $F_N$ and $P\sin\theta$ pointing upward and $mg$ pointing downward. As they have to balance one another, $$F_N+P\sin\theta=mg$$ $$F_N=150-P\sin\theta$$
Therefore, $f_2=(150-P\sin\theta)\mu_k$
The exercise states that $f_1=2f_2$, so $$150 = 2(150-P\sin\theta)$$ $$2P\sin\theta=300-150=150$$ $$\sin\theta=\frac{150}{2P}=\frac{150}{2\times110}=0.682$$ $$\theta=43^o$$
