Here's my answer:
The initial position of the king does not matter.
First of all, let's reformulate the problem.
Promoting the pawn takes 5 moves, and it will always be promoted on the same square.
So we want to find how many moves the queen will take to stalemate the king for every possible black king position in this position : Q7/8/6p1/6pb/5prp/5PpN/6Pb/4K2R w K - 0 1
Moreover, there are 11 of these squares that the king can access in 5 moves, regardless of its original position : c3, d3, e3, c4, d4, c5, e5, f5, d6, e6 and f6.
Therefore if one of these square have a 'stalemate time' higher (or equal) than every single square in this list : h6, h7, g7, f7, e7, d7, c1, c2, c7,b1, b2, b3, b4, b5, b6 (squares that you can only access in 5 moves from certain initial king positions), then the initial king position doesn't matter.
This is exactly what happens ; the b6 square takes 8 moves to stalemate and it's the maximum of list 2. But the d4 square, that we can acces in 5 moves from all the 10 king positions, also takes 8 moves to stalemate.
We end up with the following string of moves, supposing the king started of b2 (but we can also make it start on a1, b1, c1, c2, h6, h7, h8, g7 or g8) :
1. a4 Kb3 2. a5 Kc3 3. a6 Kc4 4. a7 Kc3 5. a8=Q Kd4 6. Qa5 Kc4 7. Qe5 Kb4 8. Qd5 Ka4 9. Qb7 Ka5 10. Qb3 Ka6 11. Qb4 Ka7 12. Qb5 Ka8 13. Qb6 Bg1 14. Nxg1 h3 15. Nxh3 Rh4 16. O-O
But I might be wrong.