One way to solve it: first put all $nk$ items in order: there are $(nk)!$ ways. Now chop them into blocks of $k$ and you have a partition. Each block can be reordered in $k!$ ways, then the blocks can be put in order in $n!$ ways. In all, we have $\frac{(nk)!}{(k!)^nn!}$ ways to make the partition.
Another way is to say there are ${nk \choose k}$ ways to get the first partition, ${(n-1)k \choose k}$ to get the second, and so on. Again you can choose the partitions in $n!$ orders. This gets you the same answer.
I’ll briefly sketch a bijection between these diagrams on $n$ points and valid strings of $n$ pairs of parentheses. The sketch will be a bit rough, as I’m going to have to shut down shortly and will explain it largely by example.
Start with one of your diagrams. Each point represents a matched pair ()
of parentheses. If $k<\ell$, and there is an arc from point $k$ to point $\ell$, then all of the pairs corresponding to points $k+1,\ldots,\ell$ are inside pair $k$. If this is a maximal arc, all pairs $j$ with $j<k$ are to the left of pair $k$, and all pairs $j$ with $\ell<j$ are to the right of pair $j$. If it is maximal within a larger arc from $p$ to $q$, all pairs $j$ with $p<j<k$ are to the left of pair $k$ within pair $p$, and all pairs $j$ with $\ell<j<q$ are to the right of pair $j$ within pair $p$.
These rules, applied recursively if necessary, suffice to convert your diagrams to valid parenthesis strings. For $n=4$, for instance, your diagrams, in order from the top down, correspond to the following strings, in which each pair of matched parentheses is labelled with the number of the corresponding point:
$$\begin{align*}
&(_1)_1(_2)_2(_3)_3(_4)_4\\
&(_1(_2)_2)_1(_3)_3(_4)_4\\
&(_1(_2)_2(_3)_3)_1(_4)_4\\
&(_1(_2)_2(_3)_3(_4)_4)_1\\
&(_1)_1(_2(_3)_3)_2(_4)_4\\
&(_1)_1(_2(_3)_3(_4)_4)_2\\
&(_1)_1(_2)_2(_3(_4)_4)_3\\
&(_1(_2(_3)_3)_2)_1(_4)_4\\
&(_1(_2(_3)_3(_4)_4)_2)_1\\
&(_1(_2)_2)_1(_3(_4)_4)_3\\
&(_1(_2)_2(_3(_4)_4)_3)_1\\
&(_1(_2(_3)_3)_2(_4)_4)_1\\
&(_1)_1(_2(_3(_4)_4)_3)_2\\
&(_1(_2(_2(_4)_4)_3)_2)_1
\end{align*}$$
With a bit of thought it’s clear that the procedure produces only valid parenthesis strings.
Starting with a valid string of $n$ pairs of parentheses, you simply reverse the process to get a set of arcs on a line of $n$ points. First number the pairs from $1$ to $n$ in the order in which their left parentheses appear. For example the string (()(()))
becomes $(_1(_2)_2(_3(_4)_4)_3)_1$. There are parentheses inside pair $1$, so there will be an arc from point $1$. The top level pairs inside pair $1$ are pairs $2$ and $3$, so the arc from point $1$ will be to point $3$, corresponding to the rightmost of these pairs. Pair $4$ is inside pair $3$, so there will be an arc from point $3$ to point $4$. The resulting graph is the fourth one up from the bottom in your list.
Again, a bit of thought shows that no crossing arcs are created by this process.
Best Answer
The Catalan numbers are defined by the recurrence \begin{align*} C_0 &= 1 \\ C_{n+1} &= \sum_{j=0}^n C_j C_{n-j} \text{ for } n \ge 0 \end{align*} Therefore, if $f(n)$ is the number of non-cross partitions of $\{1,2,\ldots, n\}$, we just need to show that $f(n)$ satisfies the above.
There is one non-cross partition of the empty set (the partition with no parts). So $f(0) = 1$.
Now, consider a non-cross partition of $\{1,2,3,\ldots,n+1\}$. To count how many such partitions there are, we do casework on the minimum value of the set in the partition containing $n+1$ (call this set $S$). If $S$ has minimum value equal to $k$, then:
$k$ can be anything between $1$ and $n+1$, so we have $$ f(n+1) = \sum_{k=1}^{n+1} f(k-1)f(n+1-k) = \sum_{j=0}^n f(j)f(n-j) $$
Thus $f$ and $C$ satisfy the same recurrence and initial value, so (by induction, if you like) $f(n) = C_n$ for all $n$.