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If the player is immediately to the left of the dealer, I think the answer is:

52 x (52-p) x (52-2p) x ... x (52-(n-1)p),

and if the player is the kth player, I think the answer is:

(52-k+1) x (52-p-k+1) x (52-2p-k+1) x ... x (52-(n-1)p-k+1).

Am I right?

I'm assuming that the pack has 52 cards, and that dealing starts immediately to the left of the dealer and goes round clockwise.

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There are various issues here, such as whether the order in which the cards arrive matter. If not, you might want to divide your results by $1 \times 2 \times 3 \times \cdots \times p$ – Henry Mar 27 '11 at 8:36
It's the same for all players (assuming that $n$ is a multiple of $k$). – TonyK Mar 27 '11 at 9:03

There should be no difference between the number of hands dealt to the first player or the $k$th player. To see this, imagine that you deal out the cards, then before you look at them, you force the first player to switch hands with the $k$th player. This changes nothing. There is no hand that becomes available to the $k$th player or becomes unavailable. The sets of hands are the same, and so are their counts.

Similarly, the number of players at the table does not affect the possible hands dealt to the first player. You can imagine dealing $n$ cards to the first player, then $n$ cards to the second player, etc. Whether you stop after $2$ players or $10$ does not change the hand dealt to the first player.

The number of possible hands with $n$ cards where order does not matter is

$${52 \choose n} = \frac{52 \times 51 \times ... \times (52-n+1)}{n \times (n-1) \times ... \times 1}.$$

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