Answer
The value of $\theta$ in question is $67.7^o$.
Work Step by Step
Let's consider the horizontal forces acting on the box:
- the horizontal pushing force $(P\cos\theta)$ in $+x$ direction
- the kinetic frictional force $(f_k)$ in $-x$ direction
$$\sum F_x=P\cos\theta-f_k$$
We are looking for $\theta$ so that the box cannot be moved. In other words, since there is no acceleration, $\sum F_x=0$ or the forces cancel each other out.
$$P\cos\theta=f_k$$ $$P\cos\theta=\mu_kF_N$$
The normal force $F_N$ is pretty complicated here. Besides the weight of the box, the pushing force $P$ also has a vertical component $(P\sin\theta)$ that acts in the same direction as the weight of the box. In short, $$F_N=mg+P\sin\theta$$
However, in this special case, we consider $P$ of unlimited force. No matter how large $mg$ is, $P$ can be a lot larger to make $mg$ negligible. So, we can cancel $mg$ out here. $$F_N=P\sin\theta$$
Therefore, $$P\cos\theta=\mu_k(P\sin\theta)$$ $$\cos\theta=\mu_k\sin\theta$$ $$\frac{\sin\theta}{\cos\theta}=\tan\theta=\frac{1}{\mu_k}=2.439$$ $$\theta=67.7^o$$
