Consider M6/7. These are biased with fixed gate voltage, thus they form a cascode current sink. The current is quite reasonably constant, until drain voltage drops so low that saturation is reached. (Note I'm referring to voltage saturation here.)
If there were no diode attached, that would mean M3 has no transconductance, i.e., its drain current is insensitive to gate voltage, except at low gate voltages where the current sink saturates.
M4/5/8/9 are a cascode current mirror, so the output gets whatever current M3 passes, inverted (and subtracted from the constant bias that M10/11 draw).
So the key seems to be the diode current tapping into M3 source.
We can rearrange the circuit, thinking of M3 and the diode acting as a differential pair, i.e. a current-steering function. The diode current adds to I(M6), though, not actually steering (diverting?) current but increasing it; it would be as if we could drive a diff pair with an imaginary input that causes it to deliver more current than it is biased to.

So the interpretation changes again, and M3 is simply a biased cascode: it takes the diode current, plus a bias current, and communicates it to the mirror.
What is M1-3 doing?
Making the cascode more stable, i.e. reducing the change in diode voltage. M2 is biased around one Vgs (about Vb4, give or take Vb2 and relative geometry), so M3's source must be at this potential, and M3's gate must be at about 2*Vgs. (Evidently Vdd > 3*Vgs is required here.)
The impedance of the gain node (M1/2 drain) may not be very high (as suggested by the need for cascoded sources elsewhere in the circuit), and without device characteristics, I can't calculate what this will be; but in any case, the gain acts to reduce M3's source impedance (~ 1/gm) by the gain ratio. I would guess at DC, the diode sees a load impedance of a few ohms, or tens of ohms (maybe hundreds if this is small transistors in the middle of a chip?). This keeps the photogain stable, but more important is high frequency response.
At high frequencies, capacitances (especially Cgs and Cgd) dominate, and the loop gain (of M1-3) is reduced; at high frequencies, also, the diode's junction capacitance has considerable admittance, shunting its photocurrent. This will shelve around where M2's gain drops by, say, half or more.
The slight bias also helps reduce the diode's capacitance by reverse biasing it. Though this isn't a huge difference, being maybe a volt. (It's best to run diodes near their reverse bias rating, to take full advantage of the C(V) curve. Certain diodes can be driven even further, increasing gain exponentially -- avalanche photodiodes.) But it's still better than, say, a voltage amplifier sensing an unbiased diode (i.e., photovoltaic mode).
Another advantage for reverse bias, even if it's not reducing capacitance much, is it sweeps charges out of the junction faster than recombination alone does.
This is a high-gain open-loop amplifier. Is there any additional circuitry that you did not trace?
Here is a photodiode amplifier used to charge an integration capacitor via analog switch (not shown).
Yeah, this -- since the output is simply a direct complement of the input current (a current follower, as it were: whereas a voltage follower has ~infinite input and ~0 output impedance, this has ~0 input and ~infinite output impedance), it would be an excellent pairing for anything current-mode, like an integration capacitor, or a voltage follower and feedback network.
We should probably assume that the load is a modest impedance centered around mid-supply, so that the outputs aren't saturating and the amplifier amplifies, so in the absence of anything else, assuming a feedback network would also be fair.

Incidentally, it's a current follower if the mirror has equal transistors, but some gain can be had here by making the output side a bit larger. I'm not sure by what ratio is acceptable, which is going to depend on bandwidth of course; I would guess 10-50x would be fine.
Tim