We have this theorem: Let $ G $ be a non-trivial finite supersoluble group. Then $ G $ has a normal series $ 1 = G_{0} <G_{1} < \cdots < G_{s} = G $, such that for every $ 1 \leq i \leq s $, every factor $ G_{i}/G_{i-1} $ is cyclic of order prime.
Now show if $ G $ is supersoluble then $ G $ has a chief factor such that every chief factor of order $ p $. For proof, by the above theorem $ G $ has a normal series $ G_{0} < G_{1} < \cdots \ G_{s} = G $ that every factor $ G_{i}/G_{i-1} $ is cyclic of order prime. It is sufficient show this normal series is the chief series. But how ?
Hint: every cyclic group $C$ has a series $1 < C_1 < ... < C_k = C$ such that every $C_i$ is characteristic in $C$. So if $C$ is a normal subgroup of a group $H$ then every $C_i$ is normal in $H$.