I do not understand the following:
The equation of a particular parabola is:
$$(y−23) = -\dfrac{1}{16}\!\!\!\!\!(x+3)^2$$
Given the equation of a parabola is - $(y−y_1)= a(x−x_1\!\!)^{2}$ - the number in the denominator of a (i.e. $-\dfrac{1}{16}$) is twice the distance between the directrix to the focus.
On
$(y-23) = -\frac{1}{16}(x+3)^2$ is a parabola that is congruent to $y=-\frac{1}{16}x^2$, which is much easier to analyse.
The focus of the latter parabola is at $F(0,-p)$, and the closest point on the directix from an arbitrary point $P(x,-\frac{1}{16}x^2)$ is $D(x,p)$. Of course, the vertex is at V(0, 0).
Note that the distance from $P$ to $D$ is identical to the distance from $P$ to $F$ (and hence the squares of the distances are also equal). Solve for $p$.
The distance from the focus to directix will be $2p$.
Equation of a standard downward facing parabola is $(x-x_1)^2=-4a(y-y_1),$ and in that parabola the perpendicular distance from the focus to the directrix is $2a$,
In your case, $(x_1,y_1)=(-3,23)$ and $4a=16$ which is double of $2a=8.$