$\Delta_\text r G=RT\ln \frac{Q}{K}$
If we switch $K_p$ from $K_c$, accordingly $Q_p$ will change to $Q_c$. Therefore, value of $\Delta_\text r G$ remains the same.
$ΔG^\circ=−RT\ln K=ΔH^\circ−TΔS^\circ$
If we switch $K_p$ from $K_c$, accordingly $\Delta H^\circ$ and $\Delta S^\circ$, which were defined at 1 atm will change their values according to $\pu{1 mol/L}$. Therefore, value of $\Delta G^\circ$ remains the same.
Now consider, $K_p=K_c(RT)^{\Delta n}$
$\implies K_p=K_c(0.082\times T)^{\Delta n}$ $\pu{(atm)}^{\Delta n}$
$\implies K_p=K_c(8.314\times T)^{\Delta n}$ $\pu{(pascal)}^{\Delta n}$
Which is fine, but consider:
$ΔG^\circ=−RT\ln K_p$
$\implies ΔG^\circ=−0.082\times T\ln (K_c\times0.082\times T)^{\Delta n}$ $\pu{L.atm}$ $=−8.314\times T\ln (K_c\times0.082\times T)^{\Delta n}$ $\pu{Joule}$
and also as $\ln K_p$ is dimension-less,
$ΔG^\circ=−8.314\times T\ln (K_c\times8.314\times T)^{\Delta n}$ $\pu{Joule}$
Why are there two different values of change in standard Gibbs energy? What did I miss?