If 3 is a root of the quadratic equation

It is given that 3 is a root of the quadratic equation $x^{2}-x+k=0$.

$\therefore(3)^{2}-3+k=0$

$\Rightarrow k+6=0$

$\Rightarrow k=-6$

The roots of the equation $x^{2}+2 k x+\left(k^{2}+2 k+p\right)=0$ are equal.

$\therefore D=0$

$\Rightarrow(2 k)^{2}-4 \times 1 \times\left(k^{2}+2 k+p\right)=0$

$\Rightarrow 4 k^{2}-4 k^{2}-8 k-4 p=0$

$\Rightarrow-8 k-4 p=0$

$\Rightarrow p=\frac{8 k}{-4}=-2 k$

$\Rightarrow p=-2 \times(-6)=12$

Hence, the value of p is 12.