Pressure at which graphite and diamond are in equilibrium
For each phase (graphite or diamond) you can write

$$ \mu = \mu^\circ+\int_{P^\circ}^{P} V_mdP $$
at constant T.

We are asked to find the pressure $P=P_{eq}$ at which carbon coexists in the two phases (the Gibbs free energy is equal in both) so that 

$$ \mu(diamond) = \mu(graphite)$$

This leads to

$$\Delta \mu^{\circ}=-\int_{P^\circ}^{P_{eq}}\Delta V_mdP$$

and, since $\Delta \mu^{\circ}= \Delta_f G_m^{\circ}$, ultimately to the expression you provided:

$$\Delta_f G_m^{\circ}=-\int_{P^\circ}^{P_{eq}}\Delta V_mdP$$

The approximation you are allowed to make at this point is that the solids are incompressible, such that their $V_m$ are constant with change of pressure. It follows that 

$$\Delta_f G_m^{\circ}=-\Delta V_m\int_{P^\circ}^{P_{eq}}dP=-\Delta V_m(P_{eq}-P^\circ)$$

which can be solved for $P_{eq}$:

$$P_{eq}=-\frac{\Delta_f G_m^{\circ}}{\Delta V_m}+P^\circ$$