Per unit solution gives different result
You have a problem with the sqrt(3), that's where your difference is between both calculations. Find out the value of the load impedance and check what's your current with the voltage you calculated.

Draw the full and single-line diagrams to help you better understand.

What you did in the first case is that you calculated the phase current and the impedance in a star configuration, and then applied the line-to-line voltage instead of the phase voltage.

For your first case it should be all the same, but instead of 15000 we should use 15000/sqrt(3). The value being calculated is the phase voltage.

This would give:
E = 15000/sqrt(3) + 4208,88 + 5611,842j 

E = 12869 + 5612*j = 14040 &lt; 23.56°

Remember this is the phase voltage. We multiply by sqrt(3) to get the line voltage:

E_ll = 24318 &lt; 23.56°.

Takeaway: Be careful with all the assumptions made with working in pu. In pu you normally work with single line models and voltages are line voltages, power is total three phase. Moreover all loads are assumed star connected. If you get a delta load you need to convert it to star before doing the single line and pu simplifications. Working in pu abstracts you from all this and that's why you used the wrong voltage, would've been avoided with a drawing I think.