Write $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$ and $t=e^{i\theta}$. Before measurement, the three <controlled-NOT gates> map a computational-basis component $|x,y,0\rangle$ to
$$
|x,y\mathbin\oplus x,y\mathbin\oplus x\rangle.
$$
The measured bit is therefore $b=y\mathbin\oplus x$. For $b=0$, the unnormalized state of the first qubit is
$$
\frac1{\sqrt2}(\alpha|0\rangle+t\beta|1\rangle)
=\frac1{\sqrt2}P(\theta)|\psi\rangle.
$$
For $b=1$, it is
$$
\frac1{\sqrt2}(t\alpha|0\rangle+\beta|1\rangle)
=\frac t{\sqrt2}P(-\theta)|\psi\rangle.
$$
Each branch has probability $1/2$; after normalization and removal of the irrelevant global phase $t$, the two outputs are
$$
\boxed{P(\theta)|\psi\rangle\quad\text{and}\quad
P(-\theta)|\psi\rangle}
$$
with equal probability. This is <probabilistic phase-gate injection>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 324 / 4 / b / Solution

The magic state $|A_\theta\rangle=(|0\rangle+e^{i\theta}|1\rangle)/\sqrt2$ and a short <controlled-NOT gate> circuit can apply either $P(\theta)$ or $P(-\theta)$ to an input, with a measured bit identifying the branch.

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 Incoming links: Probabilistic phase-gate injection