If he indeed swapped them by mistake, then there's no disagreement here
I did not swap units by mistake. But I believe we are in the middle of the following type of argument:
Person one says, "The sky is blue."
Person two says, "No, no, no. You are wrong. The grass is green."
I was referring to what I believe to be the customary process of load calculations. They should all be done in units of KVA. If you only have KW for some loads and don't have their KVA, you can, load by load, determine (i.e., by research or, if necessary, by intelligent guesswork) the reactive power of that load (i.e., its phase angle), and calculating the KVA value of that load. Then you can proceed by adding all KVA values.
Winnie is right in saying that two loads of 1 KVA each can actually add up to any value from 0 to 2 KVA. You will get a 0 KVA result if one is purely inductive (phase angle +90) and the other is purely capacitive (phase angle -90). But loads we usually see have phase angles not that far from each other. In addition, I can't recall ever having to include a significant capacitive load in any building I designed.
I also agree with Winnie in that adding two 1 KVA loads and getting a total of 2 KVA is no more than an approximation. Indeed, that result will likely be higher than a "true result." If you want true precision, add the KW values for all loads to get total KW, next add the KVAR values for all loads (keeping in mind that capacitive reactance is a negative value) to get total KVAR, and finally use the Pythagorean equation to get total KVA. I am certain that no previous employer of mine would have wanted to pay me to do all that extra work.
That brings me back to adding KVA to KVA to get a value of total KVA.
QED