Solution

One pass can move the money at most $k$ positions toward the driver: there are then $k-1$ people between the sender and the recipient.

If $p$ passengers hold the money, there are $p+1$ passes including the final pass to the driver. These must cover n+1 positions, so (p+1)k \ge n+1.

The answer can be computed using integer division as (n+k)/k-1. Equivalently, it is $n/k$, also using integer division.

Indeed, write n=qk+r, where 0 \le r < k. The money can pass through $q$ passengers: the first $q$ passes each cover $k$ positions, and the final pass to the driver covers r+1 \le k positions. Fewer passengers cannot suffice because $q$ passes cover at most qk \le n positions.

Complexity

The time complexity is $O(1)$ and the extra space is $O(1)$.