Let $A = \{1, 2, ..., J \}$ and $B = \{1,2, ..., K \}$.
A sequence $x^n \in A^n$ induces a probability distribution $t$ called the type where \begin{align} t(j) = \frac{1}{n} |\{x_i:x_i = j \}| \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text{ for all } j \in A \end{align}
Two sequences $x^n \in A^n$ and $y^n \in B^n$ induce a joint probability distribution on $A \times B$ called the joint type call it $s$ given by \begin{align} s(j, k) &= \frac{1}{n} |\{(x_i, y_i): (x_i, y_i) = (j, k) \}| \end{align}
We can also define a conditional type of $y^n$ given $x^n$ call it $W$ by \begin{align} W(k|j) &= \frac{s(j, k)}{t(j)} \end{align} Let's ignore the case when $t(j) = 0$. It might be irrelevant to the question I want to ask.
The number of types on $A$ is exactly equal to $$\binom{n + J - 1}{J-1}$$
Let's fix $x^n$ of type $t$. What is the number of conditional types $W$ given $x^n$ of type $t$ ? Any lower or upper bounds would help me also .
The question is when $j$ is fixed, how many types $W(\cdot|j)$ are there ? given that there are exactly $nt(j)$ element of $x$ that are equal to $j$ by a dot and bar counting argument similar to what you did, there are exactly $nt(i)+B-1\choose B-1$ such types. So now to compute the total amount of types you just need to multiply all those : \begin{align*} \prod_{j=1}^J{nt(j)+B-1\choose B-1} \end{align*}