Is it possible to determine entropy change without thermal expansion?

I was recently told by a colleague that one could "easily calculate entropy change of a solid using only Cp and isothermal compressibility constant (along with experimental parameters:Vsolid, P0, Pf, T0, and Tf) and that thermal expansion coefficient is not needed. Is this possible?

here ß is defined as isothermal compressibility and alpha is defined as thermal expansion

I have tried multiple routes to see if this is true but always end up with thermal expansion on the RHS 2nd term of dS = (Cp/T)dT + (V*alpha)dP. I was told "integrate isothermal compressibility for a V=f(P)" only which is fine and dandy for getting a pressure term grouped with dP but that doesn’t get rid of alpha (thermal expansion coefficient). The routes I have tried thus far are:

A general approach (starting with dS = MdT + NdV, invoking dV definition to sub into dV so there ends up being a dP term).

Starting with the defined enthalpy equation and making substitutions for different terms in the specific dS equation using manipulations of dV definition (subbing in for dT or dP).

And by starting with differential enthalpy and invoking differential entropy from dQ. In all cases I end up with very similar forms with the simplest being dS = (Cp/T)dT – (alphaexp(-ßP))dP.

Am I missing something? Any help is greatly appreciated, thank you 🙂