If the LEDs are blue, green, or white, then current draw is ~300mA. I = (Vsupply-Vled)/R = (4.5-3)/5.1 = 0.294, or if they're red, I = (4.5-2)/5.1 = 490mA.
You haven't shown the LEDs in your diagram. Assuming R1 is the current limiting resistor for the LEDs, then the LEDs should be in the same spot. So Vcc->Resistor->LEDs->collector of the transistor. This will work as long as the PIR produces a positive pulse when triggered; if it's inverted then you'll need to flip everything around and use a PNP transistor (or add a second transistor as an inverter).