Let’s just only consider the case for K=0, i.e. none of the 4 groups of 13 cards in the sample contains a diamond king.

Population Size, N = \binom{52}{13},
Out of these N, number of elements(group of 13) with no Diamond King M= \binom{51}{13}

Total ways to draw sample of size 4, n = \binom{N}{4}
Total ways to draw sample of size 4 with no element containing diamond king = \binom{M}{4}

Probability of proportion of samples with no element containing diamond King: P(K=0) = \binom{M}{4}/\binom{N}{4}

So I think, the calculation might be a bit different. (Calculation above is just my understanding, you can verify)
If you agree with above, you can calculate for K=1,2,3,4 and check the skewness.