Entropy and total differential
The short way is to proceed as noted in the comments. The equivalent long way is to start from the total differential for $S_m$ in terms of $p$ and $T$:

$$dS_m = \left(\frac{\partial S_m}{\partial p}\right)_Tdp + \left(\frac{\partial S_m}{\partial T}\right)_pdT$$

The second term is given by

$$ \left(\frac{\partial S_m}{\partial T}\right)_p = \frac{C_p}{T}=\frac{5R}{2T}$$

The first can be determined using the identity

$$ \left(\frac{dS_m}{dp}\right)_T = -\left(\frac{\partial V_m}{\partial T}\right)_p=-\frac{R}{p}$$

Substituting, 

$$dS_m = \left(-\frac{R}{p}\right)dp + \left(\frac{5R}{2T}\right)dT$$

It is not difficult to show that this is equivalent to the result obtained by substituting 

$$dV=\frac{R.dT}{P}-\frac{RT.dP}{P^2}$$

into the expansion in terms of $V$ and $T$ shown in the OP.