If the distance between the points A(4, p) and B(1, 0) is 5, then

(c) p = ±4 only
The given points are A(4, p) and B(1, 0) and AB = 5.

Then, $\left(x_{1}=4, y_{1}=p\right)$ and $\left(x_{2}=1, y_{2}=0\right)$

Therefore,

$A B=5$

$\Rightarrow \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}=5$

$\Rightarrow \sqrt{(1-4)^{2}+(0-p)^{2}}=5$

$\Rightarrow(-3)^{2}+(-p)^{2}=25$

$\Rightarrow 9+p^{2}=25$

$\Rightarrow p^{2}=16$

$\Rightarrow p=\pm \sqrt{16}$

$\Rightarrow p=\pm 4$