= Solution
For $f>0$ set $\tau=ft$ and $y(\tau)=p_0(t)$. The <master equation> becomes
$$
y'=-y+2p_B,\qquad p_A'=y-p_A+p_B,\qquad p_B'=p_A-3p_B.
$$
Using $p_A=1-y-p_B$ and $p_B=(y'+y)/2$ gives
$$
y''+5y'+6y=2.
$$
Initially the molecule occupies $B$, so $y(0)=0$ and $y'(0)=2$. The characteristic roots are $-2,-3$; the constant particular solution is $1/3$. Consequently
$$
y=\frac13+Ce^{-2\tau}+De^{-3\tau},\qquad C+D=-\frac13,\quad -2C-3D=2,
$$
whence
$$
\boxed{p(t)=\frac13+e^{-2ft}-\frac43e^{-3ft}.}
$$
It has the required initial slope $2f$, tends to the stationary uncharged <probability> $1/3$, and remains nonnegative. As a further check, $p_B=1/6-e^{-2ft}/2+4e^{-3ft}/3$ and $p_A=1/2-e^{-2ft}/2$ are nonnegative and sum with $p(t)$ to one. If $f=0$, no jump occurs and $p(t)=0$, also obtained directly from the boxed expression.
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