= Solution
Now $g=1$, so $c=\langle\psi|\phi\rangle$ is real and
$$
|\pm\rangle=\frac{|\psi\rangle\pm|\phi\rangle}
{\sqrt{2(1\pm c)}}.
$$
The perturbation is diagonal in this <basis>:
$$
\Delta(t)|+\rangle=2V(t)|+\rangle,
\qquad
\Delta(t)|-\rangle=0.
$$
Hence the instantaneous <eigenvalues> are $1+2V(t)$ and $-1$ with time-independent <eigenvectors>, so <Hamiltonians> at different times commute. In units $\hbar=1$, put
$$
W(t)=\int_0^tV(s)\,ds.
$$
Expanding the initial state and evolving its two components gives
$$
|\psi(t)\rangle
=\frac{\sqrt{2(1+c)}}2e^{-i[t+2W(t)]}|+\rangle
+\frac{\sqrt{2(1-c)}}2e^{it}|-\rangle.
$$
Taking its overlap with $|\phi\rangle$ and simplifying gives the exact transition probability
$$
\boxed{
\mathbb P_{\psi\to\phi}(t)
=c^2+(1-c^2)\sin^2\!\bigl(t+W(t)\bigr).}
$$
With explicit $\hbar$, each phase in the final sine is divided by $\hbar$.
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