The points $(x,y)$ are ordered orthonormal two-frames, so $P$ is the <Stiefel manifold> $V_2(\mathbb R^{n+1})$. Changing an orthonormal basis of the same plane gives the displayed free right $O(2)$-action. Since $O(2)$ is compact, the action is proper, and the quotient is the <Grassmannian> of unoriented two-planes. The <Free proper Lie-group action theorem> makes
$$
P\longrightarrow P/O(2)
$$
a principal $O(2)$-bundle.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 115 / 3 / c / Solution

{c}

If a <Lie group> acts smoothly, freely, and properly on a <smooth manifold> $P$, then the quotient $P/G$ has a unique smooth structure for which $P\to P/G$ is a <principal bundle>. An action by a compact Lie group is automatically proper.

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 Incoming links: Free proper Lie-group action theorem