= Solution
Let $\Pi_G$ be the <orthogonal projection> onto the good <vector subspace> $G$, and let the normalized input be $|\psi\rangle$. Set $p=\|\Pi_G\psi\|^2=\sin^2\theta$ with $0<\theta<\pi/2$, and define
$$
|g\rangle=\frac{\Pi_G|\psi\rangle}{\sqrt p},\qquad
|b\rangle=\frac{(I-\Pi_G)|\psi\rangle}{\sqrt{1-p}},\qquad
|\psi\rangle=\sin\theta|g\rangle+\cos\theta|b\rangle.
$$
Assuming coherent access to the two reflections, the <amplitude amplification theorem> states that
$$
Q=(2|\psi\rangle\langle\psi|-I)(I-2\Pi_G)
$$
preserves the plane spanned by $|g\rangle,|b\rangle$ and acts as a rotation, giving
$$
\boxed{Q^j|\psi\rangle=\sin((2j+1)\theta)|g\rangle+
\cos((2j+1)\theta)|b\rangle}.
$$
Hence the good-outcome probability is $\sin^2((2j+1)\theta)$. When $p$ is known, choose the nearest nonnegative integer to $\pi/(4\theta)-1/2$. The resulting angle is within $\theta$ of $\pi/2$, so the good probability is at least $\cos^2\theta=1-p$. For small $p$ this is close to one and requires $O(1/\sqrt p)$ iterations. Exact success occurs when $(2j+1)\theta=\pi/2$. Known $p$ also allows <exact amplitude amplification> by <ancilla qubit> dilution or selective phase adjustment when ordinary integer iterations would overshoot.
If $|\psi\rangle=A|0\rangle$ has a known coherent preparation, its reflection is implemented with $A$, $A^\dagger$ and a zero-state phase flip. A coherent membership test supplies the reflection about $G$. Merely possessing an unknown copy of $|\psi\rangle$ does not automatically supply its reflection. For $p=1$ the state is already good; for $p=0$ this two-reflection construction cannot generate a good component. These cases delimit the theorem's algorithmic assumptions.
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