I have a problem with your numbers.
If the maximum PD power is 12.95 watts, and the minimum PD input voltage is 37
volts, then the maximum PD current is 350 ma.
The PD input voltage varies from 37 to 50 volts with a 7 volt line drop. This
is an average input of 43.5 volts. At this input voltage the PD is only allowed
to draw 300 ma. On a short loop, the PD input could be as high as 57 volts. If
the PD draws 350 ma at this input, then the PD is consuming 19.95 watts.
The point that I am trying to make is that for a 12.95 watt PD, the maximum
current is 350 ma. This value occurs only at the longest loop with the minimum
voltage. For other cases, the PD current must be less than 350 ma.
Having a 350 ma PSE current limit allows the PD to draw at least 12.95 watts.
If the PD draws more than 350 ma, it is violating the maximum power
Hence, there is no need for 500 ma.