jmc

Notice that $p(x)=(x+2{)}^2$. The square of any real number is nonnegative, so we have $p(x)\ge 0$ for all real $x$.

Given any $y\ge 0$, we can in fact achieve $p(x)=y$ by setting $x=\sqrt{y}-2$ (or $x=-\sqrt{y}-2$), so the range of $p(x)$ is all nonnegative reals, or $\boxed{[0,\infty )}$.