As I understand your question, the problem at hand is to solve the PDE for the minority carrier concentration:
$\dfrac{\partial p}{\partial t} = D_p\dfrac{\partial^2 p}{\partial x^2} + \dfrac{p_0 - p}{\tau_p}$
on the half-line $[0, \infty)$, subject to the initial condition:
$p(x, 0) = p_0$
And the boundary conditions:
$p(0, t) = p_0 + N_m$
Where $N_{m}$ is the excess minority carrier concentration due to the photoelectric effect (reference), and
$\lim_{x \to \infty} p(x,t) = p_0.$
To solve this, considering splitting the problem into the homogeneous PDE:
$\dfrac{\partial p}{\partial t} - D_p\dfrac{\partial^2 p}{\partial x^2} + \dfrac{p}{\tau_p} = 0 $
And the inhomogeneous PDE:
$\dfrac{\partial p}{\partial t} - D_p\dfrac{\partial^2 p}{\partial x^2} + \dfrac{p}{\tau_p} = \dfrac{p_0}{\tau_p}$.
Since the system is linear, if we have have a general solution to the homogeneous equation and a particular solution to the inhomogeneous equation, then if their sum satisfies the boundary conditons it will be a solution to the original problem.
It's easy to see that a particular solution to the inhomogeneous equation is $p(x,t)_p = p_0$.
Approaching the homogeneous PDE next, we can use the Laplace transform to find a general solution that satisfies the following BC and IC:
$p(0, t) = N_m$
$p(x, 0) = 0$, and
$\lim_{x \to \infty} p(x,t) = 0$.
Taking the Laplace transform:
$\int_0^{\infty}e^{-st}f(t)dt$
In the time variable of the PDE gives us the ODE:
$s\hat{p}(x) - \mathcal{L}(p(x, 0)) = D_p\dfrac{d^2\hat{p}(x)}{dx^2} - \dfrac{\hat{p}(x)}{\tau_p} \implies $
$\dfrac{d^2\hat{p}(x)}{dx^2} = \hat{p}(x)\left(\dfrac{s}{D_p} + \dfrac{1}{D_p \tau_p}\right) = $
$\dfrac{d^2\hat{p}(x)}{dx^2} = C_n\hat{p}(x)$, where
$ C_n = \left(\dfrac{s}{D_p} + \dfrac{1}{D_p \tau_p}\right)$.
The general solution to this equation can be found by standard ODE methods and has the form:
$\hat{p}(x) = C_1e^{\sqrt{C_n}x} + C_2e^{-\sqrt{C_n}x}$.
Since we only desire solutions that decay as $x \to \infty$, the first solution is nonphysical and $C_1 = 0$.
The initial condition $p(0) = N_m\implies \hat{p}(0) = \dfrac{N_m}{s} \implies C_2 = \dfrac{N_m}{s}$,
so the solution to the ODE in the S domain is:
$\hat{p}(x) = \dfrac{N_m}{s}e^{-\sqrt{C_n}x}.$
Approaching the right hand side first, with some help from Wolfram Alpha, we find that the inverse Laplace transform of the function $P(s) = \frac{A}{s}e^{-\sqrt{s + \alpha}{x}}$ is:
$\frac{1}{2}Ae^{-\sqrt{\alpha}x}\left(\mathrm{erf}\left(\frac{2\sqrt{\alpha}t - x}{2\sqrt{t}}\right)+ e^{2\sqrt{\alpha}x}\mathrm{erfc}\left(\frac{2\sqrt{\alpha}t + x}{2\sqrt{t}}\right) + 1\right)$ for $x > 0$,
where $\mathrm{erfc}(z) = 1 - \mathrm{erf}(z)$ is the complementary error function.
Our function $\hat{p}(x)$ is of the form $\frac{A}{\beta}\frac{1}{{\frac{s}{\beta}}}e^{-\sqrt{\frac{s}{\beta} + \alpha}{x}}$, so applying the scaling theorem:
$\mathcal{L^{-1}}(P(\frac{s}{\beta})) = \beta p(\beta t)$ we finally get:
$\mathcal{L^{-1}}(\hat{p}(x)) = p(x,t)_h = \frac{1}{2}N_me^{-\sqrt{\alpha}x}\left(\mathrm{erf}\left(\frac{2\sqrt{\alpha}t\beta - x}{2\sqrt{t\beta}}\right)+ e^{2\sqrt{\alpha}x}\mathrm{erfc}\left(\frac{2\sqrt{\alpha}t\beta + x}{2\sqrt{t\beta}}\right) + 1\right)$,
where $\alpha = \frac{1}{D_p\tau_p}$, $\beta = D_p$, and the subscript "h" in $p(x,t)_h$ indicates that this is the homogeneous solution.
Summing the particular and homogeneous solution we get:
$p(x,t)_h + p(x,t)_p = p(x,t) = \frac{1}{2}N_me^{-\sqrt{\alpha}x}\left(\mathrm{erf}\left(\frac{2\sqrt{\alpha}t\beta - x}{2\sqrt{t\beta}}\right)+ e^{2\sqrt{\alpha}x}\mathrm{erfc}\left(\frac{2\sqrt{\alpha}t\beta + x}{2\sqrt{t\beta}}\right) + 1\right) + p_0$,
which satisfies the conditions of the original problem. It should be straightforward to take the limit of this expression as $t \to \infty$.
Best Answer
The concentration of electrons ($n$) and holes ($p$) in a semiconductor is determined by the Fermi-Dirac distribution function. As you correctly mentioned, the Electron-Hole Pair (EHP) generation and recombination rates are equal in thermal equilibrium. As a result, when the semiconductor is in thermal equilibrium, the concentrations $n = n_0$ and $p = p_0$ are constant (not necessarily equal). If for some reason these quantities are (say) depleted (i.e. $n < n_0$ and $p < p_0$) then the rate of EHP generation will exceed the rate of recombination until $n$ and $p$ are back to $n_0$ and $p_0$ respectively. In other words, the system returns to thermal equilibrium.
Now, in your analysis you are forgetting that there is EHP generation occurring on the n-side as well. When a EHP is generated on the n-side, the hole drifts and electron diffuses to the p-side. The current due to the electrons drifting to the n-side must cancel the current due to electrons diffusing to the p-side. Similarly, the current due to the holes diffusing to the n-side must cancel the holes drifting to the p-side. If they did not cancel each other perfectly, there would be an accumulation or depletion of holes or electrons on the n- or p-side depending on which current is dominant. When I say accumulation or depletion, I mean deviation of $n$ and $p$ from $n_0$ and $p_0$ respectively. And as I mentioned in the previous paragraph, deviation of $n$ or $p$ from $n_0$ and $p_0$ will cause the EHP generation and recombination rates to adjust such that the system returns to $n_0$ and $p_0$.