Solution:

The answer is $\mathbf{(C)}$. Notice that during the game $k\left(a_{1}+a_{2}+a_{3}\right)$ points are assigned in total, so we can write $k\left(a_{1}+a_{2}+a_{3}\right)=22+9+9=40$, from which we deduce that $k$ and $a_{1}+a_{2}+a_{3}$ are both divisors of 40. Notice moreover that since $a_{1}, a_{2}, a_{3}$ are distinct positive integers and $a_{1}>a_{2}>a_{3}$ we must have $a_{3} \geq 1, a_{2} \geq 2$ and $a_{1} \geq 3$, hence $a_{1}+a_{2}+a_{3} \geq 6$. We therefore have $k \leq \frac{40}{6}$, and listing the divisors of $40, k$ can be $1,2,4,5$. Let us now reason by cases:

- $k=1$ is clearly impossible, otherwise we would have $a_{2}=a_{3}=9$, but by hypothesis $a_{2}>a_{3}$.
- $k=2$ is impossible, otherwise we would need $2 a_{1} \geq 22$ to give Alberto enough points, but also $a_{1}<9$ because Barbara won a round, a contradiction.
- $k=4$ is impossible, indeed we would have $a_{1}+a_{2}+a_{3}=10$ and the only possible triples are $(5,3,2)$, $(5,4,1),(6,3,1),(7,2,1)$. The first two are excluded because Alberto, even with 4 wins, could not obtain 22 points. The third is excluded because Alberto, to obtain 22, would necessarily need at least 3 wins, but adding a second-place finish he would reach 21 and adding a fourth win he would reach 24. The fourth is excluded because, since Barbara won a round, her score should be at least $7+1+1+1>9$, a contradiction.
- $k=5$ is possible, indeed we have $a_{1}+a_{2}+a_{3}=8$ and the possible triples are $(4,3,1),(5,2,1)$. The first case is excluded because Alberto could not reach a score of 22 even with 5 wins. The second case, however, works, indeed Alberto to obtain 22 must necessarily have 4 wins and one second-place finish and, since by hypothesis Barbara won the first round, the ranking of the first round is necessarily Barbara-Alberto-Carlo. At this point Carlo must necessarily have come second in all the other rounds and therefore in particular in the second round.