Identify the forces acting on the box and resolve the applied force into components.

The forces acting on the box are:

• Weight ($W$) acting downwards. $W = 400$ N.
• Applied force ($P$) acting at an angle of $30.0^\circ$ below the horizontal. $P = 100$ N.
  • Horizontal component of $P$: $P_x = P \cos \theta$
  • Vertical component of $P$: $P_y = P \sin \theta$ (acting downwards)
• Normal force ($N$) acting upwards from the floor.
• Kinetic friction force ($f_k$) acting horizontally, opposite to the direction of motion.

Since the box is moving at a constant speed, the net force in both the horizontal and vertical directions is zero.

The forces in the vertical direction are the normal force ($N$) upwards, the weight ($W$) downwards, and the vertical component of the applied force ($P_y$) downwards.

$$\sum F_y = 0$$ $$N - W - P_y = 0$$ $$N = W + P_y$$

Substitute $P_y = P \sin \theta$:

$$N = W + P \sin \theta$$

Substitute the given values: $W = 400$ N, $P = 100$ N, $\theta = 30.0^\circ$.

$$N = 400 \text{ N} + (100 \text{ N}) \sin(30.0^\circ)$$ $$N = 400 \text{ N} + (100 \text{ N}) (0.5)$$ $$N = 400 \text{ N} + 50 \text{ N}$$ $$N = 450 \text{ N}$$

The forces in the horizontal direction are the horizontal component of the applied force ($P_x$) in the direction of motion and the kinetic friction force ($f_k$) opposite to the direction of motion.

$$\sum F_x = 0$$ $$P_x - f_k = 0$$ $$f_k = P_x$$

Substitute $P_x = P \cos \theta$:

$$f_k = P \cos \theta$$

Substitute the given values: $P = 100$ N, $\theta = 30.0^\circ$.

$$f_k = (100 \text{ N}) \cos(30.0^\circ)$$ $$f_k = (100 \text{ N}) (0.8660)$$ $$f_k = 86.60 \text{ N}$$

The kinetic friction force is related to the normal force by the formula:

$$f_k = \mu_k N$$

Rearrange to solve for $\mu_k$:

$$\mu_k = \frac{f_k}{N}$$

Substitute the calculated values for $f_k$ and $N$:

$$\mu_k = \frac{86.60 \text{ N}}{450 \text{ N}}$$ $$\mu_k \approx 0.1924$$

Comparing this value to the given options:
A. 0.19
B. 0.24
C. 0.40
D. 0.22

The calculated value 0.1924 is closest to 0.19.

The final answer is $\boxed{\text{A. } 0.19}$.