The correct option is option (c) $p=\pm 4$

The given points are $A(4, p)$ and $B(1,0)$ and $A B=5$.

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

$\mathrm{AB}=5$

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

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

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

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

$\Rightarrow \mathrm{p}^{2}=16$

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

$\Rightarrow \mathrm{p}=\pm 4$