put R5 with LED or you will suck more power out of the chip.
1 Meg of bias current will probably not turn the transistor on.
Were it me, I would put the ballast resistor and LED in the collector of the transistor and connect the emitter to ground. I don't know what you plan for LED current but divide it by around 50 and use that for base current. If you want 20 mA of LED current, 400 uA of base current ought to do it. I don't see any reason for connecting the LED circuit to the output but, whatever, select a resistor that will pass 400 uA at whatever the voltage is. If you decide to connect the circuit to the incoming 48 VDC, select a resistor around 100k Ohms. The value isn't critical as long as it gets the transistor into saturation. Vce should be down around 0.2V when the transistor is saturated.
Then connect the open drain PG signal to the base of the transistor.
The PG output is a small geometry MOSFET output, with a worst case
pull down Reffective of 2K. So use that in your divider calculations with
the base R for the transistor to insure you drop Vbe of transistor below
its threshold, something like .4V or less. Also factor in Vbe T dependence,
~ 2 mV / C for constant Ic.
Regards, Dana.