In order to calculate the equilibrium constant we must know the value of $\Delta_\mathrm{r}G^{\circ}$ (which is a function of $T$) so we can calculate $K$ as:
$$K = \exp\left(-\frac{\Delta_\mathrm{r}G^{\circ}}{RT}\right)$$
Now if the standard state is changed (for example from $\pu{1 atm}$ to $\pu{3 atm}$) would the equilibrium constant change? My intuition says no because we can experimentally measure $K$ which varies only with temperature. So the value of $\Delta_\mathrm{r}G^{\circ}$ must stay constant irrespective of the standard state.
What is the point then to specify the standard state?
Edit I would like to add a specific example that troubles me. Suppose we have the reaction:
$$\ce{A (g) -> B (g)}$$
that takes place in a closed container (fixed volume). We fill the container with $\ce{A}$ and after some time has passed equilibrium is established. We can predict the ratio of partial pressures of the two gases (omitting fugacity for simplicity) from the equilibrium constant $K$ which equals $\exp(-\Delta_\mathrm{r}G^{\circ}/RT)$.
What I don't understand is why if we change the standard state (picking a different pressure) must change the equilibrium constant. At least for gases it doesn't make sense because in equilibrium is their partial pressure that matters.
But for constant temperature the Gibbs free energy of the compounds must be:
$$G_i(P_i) = G(P^{\circ}) + nRT \ln \left( \frac{P_i}{P^{\circ}} \right) $$
which, as orthocresol said in the comments, shows that $\Delta_\mathrm{r}G^{\circ}$ depends on pressure. I want to find solution to that contradiction and understand the significance of the standard state.