2n3904 is being used as a switch. It controls the voltage at point B. The 10k resistor pulls point B high (to 8V) but when 2n3904 is turned "ON" then it provides a low impedance path from point B to GND. So the 10k resistor pulls point B high, but when you turn on 2n3904, it pulls point B down, and it pulls a lot harder than the resistor so it wins the tug-of-war.
R55 controls the amount of current in the base of 2n3904. You are using 2n3904 as a saturated switch, so you want the base current to be about 0.1 x the collector current (more is also OK). The collector current is 8 V / 10 k = 800 uA. So the base current should be 80 uA.
Typical base voltage in a circuit like this will be about 0.6 V.
Base current is VDD - 0.6 V / R55. So we want to set that equal to 80 uA and solve for R55.
From your schematic, VDD is 3.3 V. So:
R55 = (3.3 - 0.6) / 80 uA
R55 = 33,750 Ohms.
So you can use a 33 k resistor (33 k is a standard value). If VDD changes, or if the 10 k resistor changes, or if the switched voltage is not 8 V, then you should recalculate R55.