The mechanism given by the OP is renowned in the world of pharmacology and drug-receptor interactions, and is due to Castillo and Katz (1957). A very readable review of the history and importance of this mechanism may be found in Colquhoun (2006), published in Trends in Pharmacological Sciences. (For more great papers from the Colquhoun lab, see here)
If A is the agonist, R the vacant receptor, AR an occupied but closed form of the receptor, and AR* the occupied but open form of the receptor, the situation where an agonist must bind before the channel can open may be represented as a three state receptor model (unoccupied, occupied-closed and occupied-open) as follows:
$$\ce{A + R <=>[$k$_1][$k$_{-1}] AR <=>[\beta][\alpha] AR{*} \tag{1}}$$
We need to be more specific here, and define exactly what is meant. Let $A$ be the concentration of free agonist (ie agonist not bound to receptor), let $AR$ be the concentration of occupied closed receptor, and let ${AR^*}$ be the concentration of open occupied receptor. In addition, let $k_1$ be the second-order rate constant for the association of agonist with unoccupied receptor, and let $k_{-1}$ be the first-order rate constant for the dissociation of agonist from the occupied, closed receptor. Finally, let $\beta$ and $\alpha$ be first-order rate constants for the inter-conversion of the occupied receptor between closed and open states, respectively. Rate constants may be thought of simply as constants of proportionality arising from the law of mass action (see below).
The OP asks for a derivation of the following equation,
$$p_1^{(equil)} = \frac{({\chi}_a/{K_a})(\beta/\alpha)}{1 + ({\chi}_a/{K_a})(1+ \beta/\alpha)}\tag{2}$$
where $p_1^{equil}$ is the fraction of channels open at equilibrium, ${\chi}_a$ is the total agonist concentration, ${K_a}$ is the dissociation constant (not the association constant) for the binding of agonist to the closed receptor.
The situation is very nicely illustrated in the following diagram, taken from Colquhoun & Hawkes (1995)
Derivation
Although Eqn(2) looks daunting at first sight, the derivation is surprising simple.
Let $R_t$ be the total receptor concentration. An expression for $p_1^{equil}$ (the fraction of receptor channels open at equilibrium) may now be written as follows:
$$p_1^{(equil)} = \frac{AR^*}{R_t}\tag{3}$$
As
$$R_t = R + AR + AR^*$$
Eqn (3) may be rewritten as follows:
$$p_1^{equil} = \frac{AR^*}{R + AR + AR^*}\tag{4}$$
The problem with Eqn (3) & Eqn (4), of course, is that they are pretty close to useless. The total concentration of receptor is unlikely to be known, and obtaining reliable estimates for $AR$, $R$ and $AR^*$ present almost insurmountable experimental difficulties.
By the law of mass action, which holds that states that the rate of any reaction is proportional to the product of the reactant concentrations (and where rate constants may be considered constants of proportionality), the following may be written assuming the system described by Eqn. (1) is at equilibrium.
$$ {k_1}. A. R = k_{-1}. AR \tag{5}$$
$$ {\beta}. AR = {\alpha}AR^* \tag{6}$$
Let ${K_a}$ be the dissociation constant for the binding of agonist to the closed receptor. From Eqn (5), the following may be written.
$$ \frac{A. R}{AR} = \frac{k_{-1}}{k_1} = {K_a} \tag{7}$$
Therefore:
$$ R = \frac{{K_a} AR}{A} \tag{8}$$
An equation for the total concentration of agonist, ${\chi}_a$, may be written as follows:
$$ {\chi}_a = A + AR + AR^* \tag{9}$$
The problem with using Eqn (9) to obtain an expression for $A$ (the concentration of free agonist) is that it makes the resulting equations very difficult to solve. As regards the algebra, the situation is make a lot simpler by assuming that the total agonist concentration is equal to the free
agonist concentration, ie by assuming that Eqn (9) may be written as follows:
$$ {\chi}_a \approx A \tag{10}$$
When the concentration of agonist is very much greater than that of the receptor, this approximation is justified.
Eqn (8) may be re-written as follows:
$$ R = \frac{{K_a} AR}{{\chi}_a} \tag{11}$$
From Eqn (6), an expression for $AR^*$ may be obtained.
$$ AR^* = \frac{{\beta}. AR}{{\alpha}} \tag{12}$$
By substitution for $R$ (using Eqn 11) and $AR^*$ (using Eqn 12) in Eqn (6), followed by cancellation of the common term ($AR$), the following expression may be obtained:
$$p_1^{(equil)} = \frac{\beta/\alpha}{1 + {\beta/\alpha} + K_a/{\chi_a}}\tag{13}$$
Multiplying Eqn (13) 'above-and-below' by ${\chi_a}/K_a$ gives Eqn (2), the equation requested by the OP.
Comment
Note that when $K_a$ = $\chi_a$ and $\beta/\alpha$ >> 1, almost all of the receptors are open as $p_1^{(equil)}$ is close to 1. When $K_a$ = $\chi_a$ and $\beta/\alpha$ << 1, almost all of the receptors are closed, as $p_1^{(equil)}$ is very much less than 1.
Note also that $K_a$ is a dissociation constant and is the ratio of a first-order rate constant and a second-order rate constant, and will have dimensions of concentration.
Addendum
It may be of interest that Colquhoun's great book, Lectures on Biostatistics, is (legally) available free of charge (as a pdf) from here
DC's Improbable Science
A great site
