Answer: 49

Solution:

$P = 1 \cdot 3 \cdot 5 \cdots 199,$ so $k$ is the total number of factors of $3$ among the odd numbers up to $199.$ The odd multiples of $3$ are $3 \cdot 1, 3 \cdot 3, \ldots, 3 \cdot 65,$ and there are $33$ of them. The odd multiples of $9$ are $9 \cdot 1, \ldots, 9 \cdot 21:$ $11$ of them. The odd multiples of $27$ are $27, 81, 135, 189:$ $4$ of them. The only odd multiple of $81$ at most $199$ is $81$ itself, and there are no multiples of $243.$

Each layer contributes one additional factor of $3,$ so $k = 33 + 11 + 4 + 1 = 49.$