Note that $$x_{1}=\frac{\partial x}{\partial u^1}=(-u^2\sin(u^1),u^2\cos(u^1),p)$$ and $$x_{2}=\frac{\partial x}{\partial u^2}=(\cos(u^1),\sin(u^1),0).$$ This implies that the unit normal is given by $$x_1\times x_2=(-p\sin(u^1),p\cos(u^1),-u^2)$$ which implies that
$$n=\frac{x_1\times x_2}{\|x_1\times x_2\|}=\frac{1}{\sqrt{p^2+(u^2)^2}}(-p\sin(u^1),p\cos(u^1),-u^2).$$ Also, $$E=x_1\cdot x_1=(u^2)^2+p^2, F=x_1\cdot x_2=0, G=x_2\cdot x_2=1$$
On the other hand, we have $$x_{11}=(-u^2\cos(u^1),-u^2\sin(u^1),0),$$ $$x_{12}=x_{21}=(-\sin(u^1),\cos(u^1),0).$$ $$x_{22}=(0,0,0).$$ Therefore, we have $$e=x_{11}\cdot n=0, f=x_{12}\cdot n=p, g=x_{22}\cdot n=0.$$
Therefore, we can calculate the mean curvature using the formula (see equation (6) here): $$H=\frac{eG+gE-2fF}{2(EG-F^2)}=0$$ which shows that $M$ is minimal.