There are 39 cards left that you don't have, 5 clubs and 34 non-clubs. You need to group the cards into 3 groups of 13 cards for players P1,P2,P3.
The total number of combinations of possible hands is C(39,13)*C(26,13)*C(13,13).
The number of combinations of hands in which P1 has no clubs is C(34,13)*C(26,13)*C(13,13). Multiply this by 3 to cover # of hands where P1,P2, or P3 have no clubs. But we need to adjust for double counting.
We need to subtract for the hands in which more than one hand have no clubs. Obviously they can't all not have clubs, so we look at possibilities where 2 out of the 3 of P1,P2,P3 have no clubs.
The number of combinations where players P1 and P2 have no clubs is C(34,13)*C(21,13)*C(13,13). Multiply that by 3 to cover all combinations of 2 out of 3 players P1,P2,P3 that can have no clubs together. This result needs to be subtracted from our result from two paragraphs above to get the number of unique combinations of hands where at least one player has no clubs.
Therefore the answer is found by diving the number of combinations of hands where at least one player has no clubs by the number of combinations of possible hands.
Cancelling and combining some terms, we get...
P = (C(34,13)*C(26,13) - C(34,13)*C(21,13))*3 / (C(39,13)*C(26,13)) = 4961/14763 = .336