The problem is not very clearly defined, and I have a different interpretation from awkward's.

Based on the drawing, it seems to me that a configuration is given by a choice of "non-crossing pairings".

In other words, a configuration is a $\displaystyle 2p$-uplet $\displaystyle (i_1,\ldots,i_p,j_1,\ldots,j_p)$ such that $\displaystyle 0\leq p\leq n/2$ and:

$\displaystyle 1\leq i_1< \cdots < i_{p-1}< i_p < j_p < j_{p-1} <\cdots < j_1 \leq n$

(The bonds are between the cysteins at positions $\displaystyle i_q$ and $\displaystyle j_q$ for $\displaystyle q=1,\ldots,p$, cf. the attached drawing).

As a consequence, the number of configurations is the number of subsets of $\displaystyle \{1,\ldots,n\}$ with an even number of elements (say, $\displaystyle 2p$).

Thus the number of configurations is:

$\displaystyle N=\sum_{0\leq 2p\leq n}{n\choose 2p}$.

In fact, there are as many subsets of $\displaystyle \{1,\ldots,n\}$ with an even or an odd number of elements. This is a consequence of the following equality (using the binomial formula):

$\displaystyle \sum_{0\leq 2p\leq n}{n\choose 2p}-\sum_{0\leq 2p+1\leq n}{n\choose 2p+1}$ $\displaystyle =\sum_{0\leq k\leq n}(-1)^k {n\choose k} = (1-1)^n =0$.

We deduce that the number of even-sized subsets of $\displaystyle \{1,\ldots,n\}$ is half the number of subsets of $\displaystyle \{1,\ldots,n\}$.

As a conclusion, $\displaystyle N=\frac{1}{2}2^n = 2^{n-1}$.