 Solution

Because $k$ is algebraically closed, the smooth quadric surface $V$ has a $k$-point $p$. Choose a plane $\Pi \cong {\mathbb{P}}_k^2$ not containing $p$ and project from $p$:For a general point $q\in \Pi$, the line $\overline{pq}$ meets the degree-two surface $V$ in $p$ and exactly one further point. Sending $q$ to that second intersection is a rational map inverse to ${\pi }_p$ on dense open subsets. Therefore