If we consider that equal masses of liquid $(m)$ are taken at the same temperature then $\frac{{d\theta }}{{dt}} \propto \frac{1}{c}$

So for same rate of cooling c should be equal which is not possible because liquids are of different nature. Again from equation $(i)$

$\frac{{dT}}{{dt}} \propto \frac{{({T^4} - T_0^4)}}{{mc}}$ ==> $\frac{{d\theta }}{{dt}} \propto \frac{{({T^4} - T_0^4)}}{{V\rho \,c}}$

Now if we consider that equal volume of liquid $(V)$ are taken at the same temperature then $\frac{{dT}}{{dt}} \propto \frac{1}{{\rho \,c}}$.

So for same rate of cooling multiplication of $p× c$ for two liquid of different nature can be possible. So option $(d)$ may be correct.