Hi,
Assuming that the trigger voltage for the NPN was below 3.3V, and forgetting about resistors for now,
Never forget about resistors!
Well, I've said it over and over...
NPN goes on the low side and PNP goes on the high side.
Or to rephrase; you need the emitter to the rail, to be able to do what you think you're doing.
what voltage would the LED receive? Would the LED receive 3.3V or 5V?
Neither, but the actual voltage depends entirely on the V_f of the LED, as you have included it in the equation by putting it between emitter and the rail.
As it is, the base needs to be V_f_LED + ~0.65V above ground to start opening the transistor, so if V_f_LED > ~2.65V, the answer to your question is: less than what turns your LED on.
The emitter cannot be more than V_b - ~0.65V.
If you put the LED on the collector instead, the full voltage (here 5V) less the saturated voltage drop of the transistor will be available to kill your LED, since you don't like to include resistors and since the typical failure mode for an LED is shunted, the transistor likely follows it in the blink of an eye.
Use resistors!
