= Solution
Define interaction-picture states by $|\psi_I(t)\rangle=e^{iH_0t/\hbar}|\psi_S(t)\rangle$ and $V_I(t)=e^{iH_0t/\hbar}V(t)e^{-iH_0t/\hbar}$. Differentiation cancels the free <Hamiltonian operator> and yields
$$
\boxed{i\hbar\frac d{dt}|\psi_I(t)\rangle=\lambda V_I(t)|\psi_I(t)\rangle.}
$$
Assume the interaction is switched off sufficiently fast in the remote past for the following integral to converge. Starting in $|n\rangle$, the first iteration of the <integral equation> gives
$$
\langle m|\psi_I(t)\rangle=-\frac{i\lambda}{\hbar}\int_{-\infty}^t\langle m|V(t')|n\rangle e^{i(E_m-E_n)t'/\hbar}\,dt'+O(\lambda^2)
$$
for $m\ne n$. The picture change only multiplies the measured amplitude by a phase. Squaring therefore gives
$$
\boxed{P_{n\to m}(t)=\frac{\lambda^2}{\hbar^2}\left|\int_{-\infty}^t\langle m|V(t')|n\rangle e^{i(E_m-E_n)t'/\hbar}\,dt'\right|^2+O(\lambda^3).}
$$
For the <harmonic oscillator>, $|1\rangle=a^\dagger|0\rangle$ is normalized and $\langle1|(a+a^\dagger)|0\rangle=1$. The energy difference is $\hbar\omega$, and
$$
\int_{-\infty}^{\infty}e^{-p|t|}e^{i\omega t}\,dt=\frac1{p+i\omega}+\frac1{p-i\omega}=\frac{2p}{p^2+\omega^2}.
$$
Hence the lowest nonzero <transition probability> is
$$
\boxed{P_{0\to1}(\infty)=\frac{4\lambda^2p^2}{\hbar^2(p^2+\omega^2)^2}+O(\lambda^4).}
$$
The stronger remainder here follows from <parity>: every insertion of $a+a^\dagger$ reverses oscillator <parity>, so the amplitude between $|0\rangle$ and $|1\rangle$ has only odd powers of $\lambda$. The general displayed $O(\lambda^3)$ estimate remains valid.
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