Show the intersection of a nonidentity normal subgroup and the center of P is not trivial P is p-group and M is a nontrivial normal subgroup of P. Show the intersection of M and the center of P is nontrivial. 
By the class equation, I proved that Z(P)is not 1. Then, how do prove I the intersection of M and the center of P is not 1 or empty?  
Thank you very much for your time...
 A: $M$, being normal, is the union of conjugacy classes (with respect to $P$), meaning a conjugacy class lies completely in $M$ or is disjoint from $M$. Since the size of a conjugacy class is either $1$ or a multiple of $p$ the number of singleton classes in $M$ is a multiple of $p$ ($M$ is also a p-group), moreover it is not $0$ since the class of the identity is one of them. So there are at least $p$ singleton classes in M. But these singletons are central elements in P, which proves the assertion. With thanks to @Myself for his comment. 
A: The Class Equation for normal subgroups reads (for details, see here):
$$|M|=|M \cap Z(P)|+\sum_{m \in \{Orbits \space rep's\}}\frac{|P|}{|C_P(m)|} \tag 1$$
where:


*

*$C_P(m)$ is the centralizer of $m$ in $P$;

*"$Orbits$" (capital "O") are the conjugacy orbits in $M$ of size greater than $1$.


Now, by hyphothesis, $|M|$ is some power of $p$; moreover, $m \in \{Orbits \space rep's\} \Rightarrow C_P(m)\lneq G \Rightarrow$ $|P|/|C_P(m)|$ terms in the sum in $(1)$ are also powers of $p$; but then $|M \cap Z(P)|$ must be divisible by $p$, whence $|M \cap Z(P)|\ne 1$ and finally $M \cap Z(P) \ne \{e\}$.
