Suppose $X_1,\ldots,X_n$ are independent identically distributed random variables and
$$
\Pr(X_1 = (0,0,0,\ldots0,0,\underset{\uparrow}{1},0,0,\ldots,0,0,0)) = p_i
$$
where there are $k$ components and the single "$1$" is the $i$th component, for $i=1,\ldots,k$.
Suppose $c_1+\cdots+c_n = n$, and ask what is
$$
\Pr((X_1+\cdots+X_n)=(c_1,\ldots,c_n)).
$$
The vector $(c_1,\ldots,c_n)$ is a sum of $c_1$ terms equal to $(1,0,0,0,\ldots,0)$, then $c_2$ terms equal to $(0,1,0,0,\ldots,0)$, and so on. The probability of getting any particular sequence of $c_1$ terms equal to $(1,0,0,0,\ldots,0)$, then $c_2$ terms equal to $(0,1,0,0,\ldots,0)$, and so on, is $p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k}$. So the probability we seek is
$$
(p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k}) + (p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k}) + \cdots + (p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k}),
$$
where the number of terms is the number of distinguishable orders in which we can list $c_1$ copies of $(1,0,0,0,\ldots,0)$, $c_2$ copies of $(0,1,0,0,\ldots,0)$, and so on. That is a combinatorial problem, whose solution is $\dbinom{n}{c_1,c_2,\ldots,c_k}$. Hence the probability we seek is
$$
\binom{n}{c_1,c_2,\ldots,c_k} p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k},
$$
so there we have the multinomial distribution.
Best Answer
Well, let's review a few helpful definitions that will clarify how to calculate the Kullback-Leibler divergence here.
By definition the summation of the parameters of the mutlinomial distribution is 1; i.e., $$\sum_{m=1}^k\theta_m=1$$,
where $\theta_m$ is the probability of the $m^{th}$ outcome occuring.
The probability mass function (PMF) of the multinomial distribution is $$q(x)=\frac{n!}{\Pi_{m=1}^k(x_m!)}\Pi_{m=1}^k\theta_m^{x_m},$$ where $n$ is the total number of independent experiments executed such that $$\sum_{m=1}^kx_m=n$$.
Now let's also consider another multinomial distribution $p(x)$ as $$p(x)=\frac{n!}{\Pi_{m=1}^k(x_m!)}\Pi_{m=1}^k\left(\frac{1}{k}\right)^{x_m}=\frac{n!}{\Pi_{m=1}^k(x_m!)}\left(\frac{1}{k}\right)^{n}.$$
The resultant Kullback-Leibler divergence may then be calculated in a variety of equivalent statements $$D_{KL}(p(x)||q(x))= \sum_{m=1}^k \left(\frac{1}{k}\log\left(\frac{\frac{1}{k}}{\theta_m}\right)\right)=-\sum_{m=1}^k \left(\frac{1}{k}\log\left(k\theta_m\right)\right)=-\frac{1}{k}\log\left(k^k\Pi_{m=1}^k\theta_m\right)\\=-\frac{1}{k}\left(k\log(k)+\log(\Pi_{m=1}^k\theta_m)\right)=-\log(k)-\frac{1}{k}\sum_{m=1}^k\log(\theta_m)=\log\left(\frac{1}{k}\right)-\sum_{m=1}^k\frac{1}{k}log\left(\theta_m\right)=\sum_{m=1}^k\frac{1}{k}\log\left(\frac{1}{k}\right)-\sum_{m=1}^k\frac{1}{k}log\left(\theta_m\right)=-H\left(p(x)\right)-\mathbb{E}_{p(x)}[log(q(x))]\\=\mathbb{E}_{p(x)}\left[\log\left(\frac{1}{q(x)}\right)\right]-H(p(x)).$$ Notice that no vectors are required to calculate the Kullback-Leibler divergence between two multinomial distributions with the same number of categories.