There are several articles on estimating the solid angle of a circular disk in the physics literature. See the articles cited by the Wikipedia article on solid angles, including Gardner, R. P.; Verghese, K. (1971). "On the solid angle subtended by a circular disk". Nucl. Instr. Meth. 93 (1). pp. 163–167. That paper uses cones over approximating polygons with the same area as the circle to approximate the solid angle. Other papers mention expressing the solid angle in terms of elliptic integrals, which seems to have been considered well-known in the 1950s but not ideal for producing numerical approximations at that time. However, modern CASs can handle worse numerical integrals.
Let the point on the hemisphere be $P$. Let the equatorial disk be $D$.
Consider the right circular cones whose apex is $P$ whose axes are perpendicular to the plane of $D$. The solid angle of a cone with apex angle $2\phi$ is $2\pi (1-\cos \phi)$. The derivative of this is $2 \pi \sin \phi$. Consider the intersection of the surface of the right circular cone of apex angle $2\phi$ with the cone over $D$. This is a cone over an arc of angle $f(\phi)$ radians. The solid angle of the cone over $D$ is thus
$$\int_\theta f(\phi) \sin \phi d \phi.$$
Another way to state that is to compute the solid angle in spherical coordinates centered at $P$, which I why I used $\phi$ to denote half of the apex angle.
For $\phi \le \phi_0$, $f(\phi) = 2 \pi$, and for $\phi \ge \phi_1$, $f(\phi)=0$. Between these, we want the angle of the intersection of a circle with $D$ so that the circle intersects the boundary of $D$.
Let the distance between the perpendicular bisector and the center of $D$ be $a$. Let the distance between $P$ and the plane of $D$ be $h$. The circle $C$ has radius $r = h \tan \phi$ and is centered at a point of distance $a$ from the center of $D$. The intersection between $C$ and the boundary of $D$ occurs at angle $\theta = f(\phi)/2$ where
$$\begin{eqnarray} (r \sin \theta )^2 + (r \cos \theta - a)^2 &=& 1 \newline -2ra \cos \theta &=& 1-a^2-r^2 \newline \theta &=& \arccos \left( \frac{1-a^2-r^2}{-2ra} \right) \newline f(\phi) &=& 2 \arccos \left( \frac{1-a^2-r^2}{-2ra} \right) \newline f(\phi) &=& 2 \arccos \left( \frac{1 -a^2 - h^2 \tan^2 \phi}{-2 h a \tan \phi} \right).\end{eqnarray}$$
$\phi_0$ is the value of $\phi$ at which $C$ starts to intersect the boundary of $D$, when $a+r=1$, $a+h \tan \phi_0 = 1$, so $\phi_0 = \arctan \frac{1-a}{h}$. Similarly, $\phi_1$ satisfies $r-a=1$ so $\phi_1 = \arctan \frac{1+a}{h}$.
So, the solid angle is
$$\begin{eqnarray}\int_0^{\phi_1} f(\phi) \sin \phi d \phi =& 2\pi\left(1-\cos \arctan \left((1-a)/h\right)\right) \newline &+ \int_{\arctan (1-a)/h}^{\arctan(1+a)/h} 2 \arccos \left( \frac{1 -a^2 - h^2 \tan^2 \phi}{-2 h a \tan \phi} \right) \sin \phi d \phi.\end{eqnarray}$$
This can be simplified slightly, but this can be used in a CAS. For example, we can choose $h=3/5, a=4/5$, and Mathematica resports that the solid angle is about $2.11602$, which is greater than $(2-\sqrt 2)\pi = 1.84033.$ Here is a plot from Mathematica of the solid angles as a function of the displacement from the axis $a$:
So, numerical calculations indicate that this is not constant.