= Solution
Define the <interaction picture> by
$$
|\psi_I(t)\rangle=e^{iH_0t/\hbar}|\psi_S(t)\rangle,\qquad
V_I(t)=e^{iH_0t/\hbar}V(t)e^{-iH_0t/\hbar}.
$$
Differentiating and using the <Schrödinger equation> cancels the $H_0$ terms, leaving
$$
\boxed{i\hbar\frac d{dt}|\psi_I(t)\rangle=\lambda V_I(t)|\psi_I(t)\rangle}.
$$
Its integral equation, iterated once from the initial $|a\rangle$, gives $|\psi_I(t)\rangle=|a\rangle-(i\lambda/\hbar)\int_0^tV_I(t')|a\rangle\,dt'+O(\lambda^2)$.
Since $a,b$ are <orthogonal> <energy eigenstates> with distinct <energies>, the zeroth-order transition amplitude is zero, while
$$
\langle b|V_I(t')|a\rangle
=e^{i(E_b-E_a)t'/\hbar}\langle b|V(t')|a\rangle.
$$
Returning to the Schrodinger picture changes this component only by a unit-modulus phase. Squaring the first-order amplitude therefore gives
$$
\boxed{P_{a\to b}(t)=\frac{\lambda^2}{\hbar^2}
\left|\int_0^t\langle b|V(t')|a\rangle
e^{i(E_b-E_a)t'/\hbar}\,dt'\right|^2+O(\lambda^3)}.
$$
For exponential switching, put $\Delta=E_b-E_a$. The integral is
$$
\langle b|W|a\rangle\,
\frac{\hbar[1-e^{(-\mu+i\Delta)t/\hbar}]}{\mu-i\Delta}.
$$
Because $\mu>0$, its exponential vanishes at late time, yielding
$$
\boxed{P_{a\to b}(\infty)\simeq
\frac{\lambda^2|\langle b|W|a\rangle|^2}{\mu^2+\Delta^2}}.
$$
This is the leading weak-perturbation result, not an exact <probability> formula at arbitrary coupling.
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