The number of $K$-permutations of the numtiset $\{ 1^D, 2^D, \ldots, N^D \}$ is
$$\sum_{j_1+j+2+\dots+j_N=K\atop 0 \leq j_i \leq D} \binom{K}{j_1,j_2,\dots,j_N}.$$
(summands here are multinomial coefficients)

Alternatively, denoting by $m_i$ the number of $j$'s equal $i$, we get a formula as the sum over restricted partitions:
$$\sum_{0m_0+1m_1 + \dots + Dm_D = K\atop m_0 + m_1 + \dots + m_D = N,\quad m_i\geq 0} \binom{N}{m_0,\dots,m_D} \frac{K!}{1!^{m_1} 2!^{m_2} \cdots D!^{m_D}}$$

For the example with $N=9$, $D=3$, $K=4$, the latter formula consists of four summands and gives:
$$\binom{9}{5,4} \frac{4!}{1!^4} + \binom{9}{6,2,1}\frac{4!}{1!^2 2!^1} + \binom{9}{7,1,1}\frac{4!}{1!^1 3!^1} + \binom{9}{7,2}\frac{4!}{2!^2}$$
$$ = 3024 + 3024 + 288 + 216 = 6552$$
as expected.