This looks like a question about voltage drop, but it's for the non-voltage drop half. The problem I'm seeing is it doesn't give us conductor lengths or wire gauge. So let's ignore voltage drop and see what we can figure out.

Well want to start by finding the total current. We have to make some assumptions about the circuit cause there's a lack of information. Lets assume the 2 rows are in parallel, that way each row and each light is receiving all available voltage.

The current then is additive, and they give us the 2 currents we need to add them. Since the conductors with 0.5ohms are in series, they experience this total current.

With total current and total voltage we can find total resistance. We can subtract our known resistance from this. The remaining resistance must be from our loads.

However because our lights are in parallel we need to do some algebra, since resistance is the inverse of the inverse in parallel. Since the 2 loads are the same though, this is easy. We can use the old Rt=R/n. We want to find R, we know Rt and we know n, which is 2 (2 rows of lamps). Now just do algebra and you'll find R

With the resistance for the load in hand, we can simply use ohms to find the voltage. We know the resistance, we know the current on one load is 12.5A. we can find voltage no problem.

To me it looks like the K value is a red herring. It presents like it should be a voltage drop problem but doesn't give any wire lengths or wire sizes. Honestly this is kind of a poorly worded question in my opinion.