Turn on LED strip via PIR using NPN transistor
Add a base resistor of 100 &Omega; or so. Your present design overloads the PIR because it has no current limiting and the base-emitter junction behaves like a diode.

Your question is missing the current requirements of the LED strip. If you edit and add that in we can see if the transistor is adequately rated.





Figure 1. The transistor should be capable of 2 A. You may need heatsinking.


  Ah okay. LED strip requires 1A. Added 100 ohm resistor which limits LED strip to 2.5V. If I need to have 5V then I need 1 ohm resistor right? But then base draws 2.4A ?


No, read it again. Add the resistor to the base. 100 &Omega; will limit the current from the microcontroller to the base to about \$ \frac {3.3 - 0.6} {100}  = 27 \ \text{mA} \$. This should be enough to drive the transistor into saturation and provide a very low voltage drop between the collector and the emitter applying almost 5 V to your LED strip.