= Solution
Apply <exact quantum phase estimation> to $U(t)$ with the supplied <eigenstate> $|\Psi\rangle$. Since
$$
U(t)|\Psi\rangle=e^{2\pi i\phi}|\Psi\rangle,
\qquad
\phi=-\frac{\lambda t}{2\pi}\pmod1,
$$
the promise that the phase has an $N$-bit representation makes an $N$-qubit control register recover $\phi$ exactly. Multiplying by $-2\pi$ modulo $2\pi$ gives $\lambda t\bmod2\pi$. Equivalently, phase estimation may be run on $U(t)^\dagger$, whose eigenphase is $\lambda t/(2\pi)$ modulo one.
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