What kind of equilibrium constant we use for Gibbs free energy and Van't Hoff equation? We know that the Gibbs free energy is related to the equilibrium constant by the following equation:
$$\Delta_\mathrm{r}G^\circ=-RT\ln K$$
We also know the Van't Hoff equation:
$$\ln\left(\frac{K_2}{K_1}\right)=\frac{-\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$

$\left(R = 8.314\ \mathrm{J~mol^{-1}~K^{-1}}\right)$

My question is: what kind of $K$ do we use for these equations? Concentration quotient ($K_c$) or pressure quotient ($K_p$)?
The first equation actually contains the definition of the standard equilibrium constant:
$$K^\circ = \exp\left\{\frac{−\Delta_r G^\circ}{R T}\right\}$$
With this definition the equilibrium constant is dimensionless. 

Under standard conditions the van't Hoff equation is
$$\frac{\mathrm{d} \ln K^\circ}{\mathrm{d}T} = \frac{\Delta H^\circ}{R T^2},$$
and therefore uses the same constant. The integrated variant is therefore already an approximation and may be correct using a different definition of the equilibrium constant.
$$\ln \left( {\frac{{K_{T_2} }}{{K_{T_1} }}} \right) =
  \frac{{\ - \Delta H^\circ }}{R}
  \left( {\frac{1}{{T_2 }} - \frac{1}{{T_1 }}} \right)$$

Now the ordinary equilibrium constant may be defined in various forms:
$$K_x = \prod_B x_B^{\nu_B}.$$

Probably one of the best representations for the standard equilibrium constant involves relative activities, for an arbitrary reaction, 
$$\ce{\nu_{A}A + \nu_{B}B -&gt; \nu_{C}C + \nu_{D}D},$$
this resolves in 
$$K^\circ = \frac{a^{\nu_{\ce{C}}}(\ce{C})\cdot{}a^{\nu_{\ce{D}}}(\ce{D})}{a^{\nu_{\ce{A}}}(\ce{A})\cdot{}a^{\nu_{\ce{B}}}(\ce{B})}.$$

The concentration is connected to the activity via
$$a(\ce{A})= \gamma_{c,\ce{A}}\cdot{}\frac{c(\ce{A})}{c^\circ},$$
where the standard concentration is $c^\circ = \pu{1 mol/L}$. 
At reasonable concentrations it is therefore fair to assume that activities can be substituted by concentrations, as
$$\lim_{c(\ce{A})\to\pu{0 mol/L}}\left(\gamma_{c,\ce{A}}\right)=1.$$
See also a very detailed answer of Philipp.

The partial pressure is connected to the activity via
$$a(\ce{A}) = 
\frac{f(\ce{A})}{p^{\circ}} = \phi_{\ce{A}} y_{\ce{A}} \frac{p}{p^{\circ}},$$
with the fugacity $f$ and the fugacity coefficient $\phi$ and the fraction occupied by the gas $y$, the total pressure $p$, as well as the standard pressure $p^\circ=\pu{1 bar}$ or traditional use of $p^\circ=\pu{1 atm}$.
For low pressures it is also fair to assume that you can rewrite the activity with the partial pressure $p(\ce{A})$, since
\begin{align}
\lim_{p\to\pu{0 bar}}\left(\phi_{\ce{A}}\right) &amp;=1, &amp;
p(\ce{A}) &amp;= y_{\ce{A}}\cdot{}p.
\end{align}

Of course concentrations and partial pressures are connected via the ideal gas
\begin{aligned}
pV\ &amp;=nRT\\
p\ &amp;\propto \text{const.} \cdot c,
\end{aligned}
and therefore it is valid to write:
$$K_c\propto \text{const.} \cdot K_p.$$
It is important to note, that the two expressions are not necessarily being equal, and can scale by powers of $(\mathcal{R}T)^{\sum{}\nu}$.

While using these equations it is always necessary to keep in mind, that there are a lot of approximations involved, so it depends very much on what you are looking for. Either use might be fine, as all these functions are related - some might lead to simple, some may lead to complicated solutions.