So for a given time interval the laser will have a mean photon number, that is the average number of photons present in a pulse that is of that interval. There will be a probability distribution around this though, Poissonian in this case. So within your exposure time you will have a probability distribution of detecting $n$ photons as $P(n)$.

I think what you're asking is how does the photon shot noise translate into electrical signal noise? You will have the responsivity of the pixel $R$ in units of Amps/Watt (Tells you how many amps of photocurrent you get for a watt of optical power) then you will have your optical power in that exposure time which is $N_{photons}hf/t_{exposure}$. Your photocurrent is just the product of the two. However $N_{photons}$ will we probabilistic, i.e. given by the probability of having $n$ photons given by $P(N_{photons})$. This gives you the probability distribution of photocurrent.

So for a given time interval the laser will have a mean photon number, that is the average number of photons present in a pulse that is of that interval. There will be a probability distribution around this though, Poissonian in this case. So within your exposure time you will have a probability distribution of detecting $n$ photons as $P(n)$.

I think what you're asking is how does the photon shot noise translate into electrical signal noise? You will have the responsivity of the pixel $R$ in units of Amps/Watt (Tells you how many amps of photocurrent you get for a watt of optical power) then you will have your optical power in that exposure time which is $N_{\text{photons}}hf/t_{\text{exposure}}$. Your photocurrent is just the product of the two. However $N_{\text{photons}}$ will we probabilistic, i.e. given by the probability of having $n$ photons given by $P(N_{\text{photons}})$. This gives you the probability distribution of photocurrent.