How is the association/ dissociation constant derived?
This looks a lot like a normal equilibrium constant. There is very little background information provided, but I assume that this formula applies to an experimental procedure in which the same total amount of the solute is dissolved into each solvent and the concentrations of free (dissociated) solute are measured. In this case that $C_1$ and $C_2$ should be the other way round. In the following, I assume that $C_1$ is the measured concentration in benzene and $C_2$ is the measured concentration in water.

In order for the formula to work we also need to assume that in benzene (or whatever the other solvent is), the solute associates almost completely.

With these assumptions the derivation of the formula is practically identical to the standard derivation of the formula for an equilibrium constant.

Let's call a single molecule of the solute A and a complex of $n$ associated molecules B. 

The association reaction then is

$$n\ce{A &lt;=&gt;[k_1][k_2] B},$$

where $k_1$ and $k_2$ are the rates of association and dissociation reaction respectively.

The speed of the dissociation reaction is proportional to the concentration of B, so we can write $k_2=b\ce{[B]}$.

The association reaction requires $n$ units of A coming together and from probability theory we can tell that the probability of that happening is proportional to the concentration of A multiplied by itself $n$ times, so we can write $k_1=a\ce{[A]}^n$.

At equilibrium the rates of association and dissociation reactions are equal, so we can write:

$$k_1=k_2 \Rightarrow a\ce{[A]}^n=b\ce{[B]} \Rightarrow \frac{b}{a}=\frac{\ce{[A]}^n}{\ce{[B]}}$$

Now we raise both sides to the power of $\frac{1}{n}$ in order to get

$$\left(\frac{b}{a}\right)^{\frac1n}=\frac{\ce{[A]}}{\ce{[B]}^{\frac1n}}$$

Now, $\ce{[A]}$ is the concantration of dissociated solvent in benzene, so we can substitute $\ce{[A]}=C_1$. On the other hand $C_2$ is the concentration of the solute in water, which is the same as the concentration of total concentration of both associated and dissociated forms in benzene, so we can write $C_2=\ce{[A]}+n\ce{[B]}\approx n\ce{[B]}$ if $\ce{[A]}$ is much smaller than $\ce{[B]}$ as we assumed in the beginning.

By substituting we can write

$$\left(\frac{b}{a}\right)^{\frac1n}=\frac{C_1}{\left(\frac1n C_2\right)^{\frac1n}} \Leftrightarrow \left(\frac{b}{na}\right)^{\frac1n}=\frac{C_1}{C_2^{\frac1n}}$$

Because $n$ is independent of $C_1$ and $C_2$, we can substitute $K=\left(\frac{b}{na}\right)^{\frac1n}$ and we get

$$K=\frac{C_1}{C_2^{\frac1n}}.$$