= Solution
Because $k$ is <algebraically closed field>[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 $p$:
$$
\pi_p:V\dashrightarrow\Pi.
$$
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
$$
\boxed{V\text{ is birational to }\mathbb P_k^2}.
$$
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