Diode clamp circuit question
I don't think that the annotations are really correct, and the schematic is not what was intended.

Anyway...

Obviously the top diode will clamp over voltages, but it will clamp them to 5.7V.
By having the lower diode, 0.7V is dropped, and so when clamped, Vout does not exceed 5V. However there is a series diode drop in the output - probably not very good.

But why?

The clue may be in the word "comparator". Bad things such as state inversion can happen when inputs to comparators and opamps go outside the max operating range. The designer may have a hard need to keep the input from exceeding VDD.

This is how you would clamp without exceeding the rails, and without having a diode drop in series:



Note that diode clamps like this are soft clamping - current will start to flow when VIN reaches 4.8-4.9V, so ADC reading for example will be wrong at the edge (4.9V) as the clamp starts to conduct.