50 Kings,14 Knights:
This is optimal but not unique, see bottom of this answer.
I think the problem is equivalent to covering every square on the board with as few knights as possible and this can be split up into black and white squares. So proving that we need at least seven knights to cover all black squares would suffice.
The above was constructed from this solution to the equivalent problem:
Proof of optimality of this:
To cover the 4 black squares of any file at least two N's are required. For the first and last files these two have to be on the same file themselves and once the file is chosen there is only one way of placing the two N's. (If 3 or 4 N's are used we will still have at least two in one of the files in reach by pigeon hole principle.) Therefore we have 4 N's in two files (b or c and f or g). In particular, they cannot contribute to covering these two files. If fewer than 7 were possible, then we would have to cover both files with the last two N's. These two would then have to sit in the file in middle between the others, leaving only the two symmetric scenarios of two N's in each of either b,d,f or c,e,g. And we can check that this is not a solution:
Note that rotating the board 90° and reapplying this argument almost necessarily leads to our 7 N solution. But there is one alternative solution. (Thanks to @Daniel Mathias for pointing that out and also that there are no other solutions):
Using this we can construct two more solutions to the original problem: