Is the equilibrium constant in the expression based on pressure or concentration?
Neither. It is exactly what is given there $K$, the standard equilibrium constant, and the equation given is its definition,
$$\Delta G^\circ=-RT\ln K.\tag{1}\label{def}$$

A general equilibrium constant for any reaction 
$$\ce{\nu_A A + \nu_B B + \nu_C C + \cdots 
   &lt;=&gt;\nu_{A'} A' + \nu_{B'} B' + \nu_{C'} C' + \cdots  }$$
is given by
$$K_x = \frac{ x_{\ce{A'}}^{\ce{\nu_{A'}}}\cdot x_{\ce{B'}}^{\ce{\nu_{B'}}}\cdot
               x_{\ce{C'}}^{\ce{\nu_{C'}}}\cdots}{
               x_{\ce{A}}^{\ce{\nu_A}}\cdot x_{\ce{B}}^{\ce{\nu_B}}\cdot
               x_{\ce{C}}^{\ce{\nu_C}}\cdots},$$
or more general, gathering all components in $\mathbb{B}=\{\ce{A, B, C, \dots, A', B', C', \dots }\}$
$$K_x = \prod_{B~\in~\mathbb{B}} x_B^{\nu_B}.\tag2$$

If you insert that into \eqref{def}, then $x$ becomes the chemical potential of the component. You can derive the link to the activity of the component (see here), and you can link the activity to concentrations, pressures, etc.

Given the right boundary conditions, e.g. ideal dilution, $K$ essentially becomes $K_c$, etc. 

$$\mu(i) \propto a(i) \propto c(i) \implies K \propto K_c\tag3$$

Obviously, this is a very sloppy generalisation, and you should study thermodynamics in more detail to fully understand the differences between the quantities used, and when to use which.