The photoelectrons are emitted with random angles as well as random energies and it's for these reasons that not all the electrons will reach the collecting plate.
Some photoelectrons will be emitted at angles that simply miss the collecting plate. How much this matters will depend on the geometry. Your diagram shows a plate spacing larger than the plate diameter and in this case you'd miss quite a lot of the emitted electrons, though I assume your diagram is just an illustration and in reality you'd use large area plates close together to reduce the effect. Anyhow, using a positive voltage will collect electrons that would otherwise be lost.
Also some photoelectrons will have low enough energies that they are scattered by air molecules before they can reach the collecting plate. A positive voltage collects electrons that would otherwise be scattered. I note you've mentioned in a comment that your system is evacuated so this wouldn't apply. However it will matter for any experiments that aren't done in a vacuum.
Finally I would challenge your diagram. That diagram appears to originate on this web page and it's just an illustration and not real experimental data. It isn't clear how big the increase in current with positive voltage actually is in a real experiment. Actual experimental data is hard to find but what I have found suggests the effect of positive voltage can range from relatively small to almost zero. The variation is presumably due to variations in how the experiment is done. If you use a large collecting plate in vacuum I would expect the effect of a positive voltage on the current to be very small.
When we graph photocurrent against the voltage applied against the electrons moving, why does the photocurrent increase when the voltage increased (until it reaches saturation current).
