Consider the two vectors to the points on the sphere,
$${\bf v}_i=(r\sin\theta_i\cos\varphi_i,r\sin\theta_i\sin\varphi_i,r\cos\theta_i)$$
with $i=1,2$. Use the dot product to get the angle $\psi$ between them:
$${\bf v}_1\cdot {\bf v}_2=r^2\left(\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2\cos\left(\varphi_1-\varphi_2\right)\right)=r^2\cos\psi.$$
Then the arclength is
$$s=r\psi=r\cos^{-1}\left(\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2\cos\left(\varphi_1-\varphi_2\right)\right).$$
Each great circle on a sphere centered at the origin corresponds to a unique plane through the origin; the intersection of two great circle arcs can be computed using the corresponding planes and linear algebra.
The procedure below assumes $B$ and $D$ do not lie on a diameter of the sphere (i.e., are distinct and not antipodal).
Synopsis: Convert $B$ and $D$ to Cartesian coordinates, and let $N$ be a (non-zero) normal vector to the plane of the black circle. If
$$
n = \frac{B \times D}{\|B \times D\|},\qquad
v = n \times B,
$$
then
$$
H = \frac{-(N \cdot v)B + (N \cdot B)v}{\|-(N \cdot v)B + (N \cdot B)v\|}.
$$
(This gives Cartesian coordinates, of course, which may need to be converted back to spherical, the details of which depend on your conventions for spherical coordinates.)
Details: Convert the spherical coordinates of $B$ and $D$ to Cartesian spatial coordinates using your favorite convention. For example, if $\theta$ denotes longitude (with the positive $x$-axis on the prime meridian) and $\phi$ latitude, then
$$
(x, y, z) = (\cos\theta \cos\phi, \sin\theta \cos\phi, \sin\phi),\qquad
-\pi < \theta \leq \pi,\ -\pi/2 \leq \phi \leq \pi/2.
\tag{1}
$$
The normalized cross product
$$
n = \frac{B \times D}{\|B \times D\|}
$$
is orthogonal to the plane $P$ containing the spherical arc $BD$. (In the preceding formula, $B$ and $D$ denote Cartesian coordinate vectors obtained from (1).) The unit vector
$$
v = n \times B
$$
lies in $P$, is orthogonal to $B$, and "points toward $D$" in the following sense: Every point of the spherical arc through $B$ and $D$ has the form
$$
B \cos t + v \sin t
\tag{2}
$$
for some $t$ with $0 < t \leq \arccos(B \cdot D) < \pi$.
The point $H$ you seek is the unique point of the form (2) lying in the plane $P'$ of your black circle. To find this point, let $N$ denote a non-zero normal vector of $P'$; the point (2) lies in $P'$ if and only if
$$
0 = N \cdot (B \cos t + v \sin t)
= (N \cdot B) \cos t + (N \cdot v) \sin t.
\tag{3}
$$
A bit of algebra (rearranging (3) and substituting into (2)) gives
$$
H = \frac{-(N \cdot v)B + (N \cdot B)v}{\|-(N \cdot v)B + (N \cdot B)v\|}.
$$
Best Answer
Consider the full great circle that your $Q$ to $R$ arc is part of. There are two points (poles) which are each 90 degrees of arc from every point on the great circle. (You can find these points by forming a 90-90-x triangle from your arc segment.) Choosing the point which is on the same hemisphere as your point $P$, there is an arc from this pole, through $P$, to the great circle. Since the length of this arc is 90 degrees, the distance from $P$ to the great circle is the complement of the distance from the pole to $P$.
Determining the distance from $P$ to the actual arc then is just a matter of determining if the intersection point is between $Q$ and $R$, and if not, determining the distances from $P$ to $Q$ and $R$ to get the minimum.
A note to anyone who objects to my use of degrees: sorry, but I use my spherical trig on the actual earth, where lat and long are still measured in degrees. And yes, I am aware it is not a true sphere, but that is a separate topic.