= Solution
Put $k=\beta F$ and $q=k\delta$. The <absorbing boundary condition for a diffusion> gives $A=j_0e^q/(Dk)$, hence
$$
p(x)=\frac{j_0}{Dk}\left[e^{k(\delta-x)}-1\right].
$$
This is nonnegative throughout the allowed interval. Normalizing the <probability density> gives
$$
1=\frac{j_0}{Dk}\left[\frac{e^q-1}{k}-\delta\right]
=\frac{j_0}{Dk^2}(e^q-1-q).
$$
Therefore
$$
\boxed{j_0=\frac{D\beta^2F^2}{e^{\beta F\delta}-\beta F\delta-1},\qquad
A=\frac{\beta F\,e^{\beta F\delta}}{e^{\beta F\delta}-\beta F\delta-1}.}
$$
Equivalently the normalized density is
$$
p(x)=\frac{k[e^{k(\delta-x)}-1]}{e^q-1-q}.
$$
The denominator is positive for $q>0$, since $e^q>1+q$. The dimensions are consistent: $j_0$ has units of inverse time, $A$ and $p$ of inverse length, and $q$ is dimensionless.
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