Will that be a constant?

Without loss of generality let $r$ be a unit (rescaling does not change trigonometric ratios) and let us introduce coordinates, such that the centre of $\Gamma = (0,0)$ , and $A = (1, 0)$ . Then we have $B = (1, -12), \ C = (1, 24)$ .

$P$ lies on the unit circle, so if its coordinates are $(-x,-y)$ , then $x^2+y^2=0$ .

$\vec{PB}=(x+1, y-12)$ ; $\vec{PA}=(x+1, y)$ and $\vec{PC}=(x+1, y+24)$ .

$\cos \angle APB = \frac{\vec{PB} \cdot \vec{PA}}{|\vec{PB}| \times |\vec{PA}|} = \frac{(x+1)^2 + y(y-12)}{d}$ , where

$d = \sqrt{((x+1)^2+(y-12)^2)((x+1)^2+y^2)} =$

$= \sqrt{(x+1)^4 + (x+1)^2 (y^2 + (y-12)^2) + y^2 (y-12)^2}$ .

Now $\cos^2 \angle APB = \frac{(x+1)^4 + y^2(y-12)^2 + 2 (x+1)^2y(y-12)}{d^2}$ , so

$\sin^2 \angle APB = 1 - \cos^2 \angle APB =$

$= \frac{d^2 - (x+1)^4 - y^2(y-12)^2 - 2 (x+1)^2y(y-12)}{d^2} =$

$= \frac{(x+1)^2 (y^2 + (y-12)^2 - 2y(y-12))}{d^2} =$

$= \frac{(x+1)^2 (y - (y - 12))^2}{d^2} = \frac{144 (x+1)^2}{d^2}$ ,

hence $\sin \angle APB = \frac{12(x+1)}{d}$ and so

$\cot \angle APB = \frac{\cos \angle APB}{\sin \angle APB} = \frac{(x+1)^2 + y(y-12)}{12(x+1)}$ .

Analogically (replacing -12 by 24 everywhere), we see that

$\cot \angle APC = \frac{(x+1)^2 + y(y+24)}{24(x+1)}$ .

Hence the fraction in questions is equal to:

$(\frac{(x+1)^2 + y(y+24)}{24(x+1)} + \frac{(x+1)^2 + y(y-12)}{12(x+1)})^{-1} =$

$=(\frac{((x+1)^2 + y(y+24) + 2(x+1)^2 + 2y(y-12) }{24(x+1)})^{-1} =$

$=\frac{24(x+1)}{3x^2 + 3y^2 + 6x + 3}$ .

Note that $x^2 + y^2 = 1$ as $P$ lies on the unit circle, hence the expression above is:

$\frac{24(x+1)}{3 + 6x + 3} = \frac{24(x+1)}{6(x+1)} = 4$ .