If −5 is a root of the quadratic equation

It is given that $-5$ is a root of the quadratic equation $2 x^{2}+p x-15=0$.

$\therefore 2(-5)^{2}+p \times(-5)-15=0$

$\Rightarrow-5 p+35=0$

$\Rightarrow p=7$

The roots of the equation $p x^{2}+p x+k=0=0$ are equal.

$\therefore D=0$

$\Rightarrow p^{2}-4 p k=0$

$\Rightarrow(7)^{2}-4 \times 7 \times k=0$

$\Rightarrow 49-28 k=0$

$\Rightarrow k=\frac{49}{28}=\frac{7}{4}$

Thus, the value of $k$ is $\frac{7}{4}$.