The correct answer is $\boxed{\text{B}}$.
The horizontal range of a projectile is the distance it travels horizontally after being launched. The maximum height of a projectile is the highest point it reaches after being launched. The angle of projection is the angle between the initial velocity vector of the projectile and the horizontal.
The horizontal range and maximum height of a projectile are equal when the angle of projection is $\boxed{45^\circ}$. This is because the horizontal range is given by the formula
$$R = \frac{v^2 \sin(2 \theta)}{g}$$
where $v$ is the initial velocity of the projectile, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity. The maximum height is given by the formula
$$H = \frac{v^2 \sin^2(\theta)}{2g}$$
When $\theta = 45^\circ$, $\sin(2 \theta) = 1$ and $\sin^2(\theta) = \frac{1}{2}$. Substituting these values into the formulas for $R$ and $H$, we get
$$R = \frac{v^2}{g}$$
$$H = \frac{v^2}{4g}$$
Therefore, the horizontal range and maximum height of a projectile are equal when the angle of projection is $45^\circ$.
The other options are incorrect because they are not angles that make the horizontal range and maximum height of a projectile equal.