Prove that for a given point on an ellipse, the sum of the distances from each focal point is constant

Note that $y=0$ gives $x=\pm5$, so $-5 \le p \le 5$ for all points $P(p,q)$ on the ellipse. Therefore:

$$ |F_1P| = \frac{1}{5}\sqrt{(4p+25)^2} = \frac{1}{5}|4p+25| = \frac{1}{5}(25+4p) \\ |F_2P| = \frac{1}{5}\sqrt{(4p-25)^2} = \frac{1}{5}|4p-25| = \color{red}{\frac{1}{5}(25-4p)} $$

The key mistake you made here is your simplification of square roots. Actually, since a square root is positive always, you should get $\vert F_1P\vert = \frac{|4p+25|}{5} $ and $ \vert F_2P\vert = \frac{|4p-25|}{5}$. However, note $ -5 \leq p \leq 5$, so $ -20 \leq 4p \leq 20$ and $\frac{|4p+25|}{5} = \frac{4p+25}{5}, \frac{|4p-25|}{5} = \frac{-4p+25}{5}$ Summing the two distances, $p$ cancels out.

Since $$e^2= a^2-b^2 = 25-9 =16\;\; \Longrightarrow \;\;e =4$$ so the point $F_1$ and $F_2$ are focuses of ellipse and by definiton of ellipse we have $$PF_1+PF_2 = constant = 2a =10$$