Let $G$ a finite group. I use notation on this reference for $O_p(G), O_{p'}(G), O_{p',p}(G)$.
Furthermore I define the $p$-residual $O^p(G)$ as the minimum normal subgroup with quotient $\frac{G}{O^p(G)}$ is a $p$-group. Now I can proof this statement (on Wikipedia):
'A group is $p$-nilpotent (i.e. it has a normal $p$-complement) if and only if it is equal to its own $p′,p$-core'.
I have proof $(\Rightarrow)$:
For hypothesis exist $K\triangleleft G$ and $P$ $p$-Sylow subgroup of $G$ with $G=PK$ and $P\cap K=1$. I know that $O^p(G)$ is generated by $p'$-elements of $G$. I have proof (omitted) that $K$ coincides with set of $p'$-element and consequentially $O^p(G)=\langle K\rangle=K$ is in particular a normal $p'$-subgroup of $G$. By definitionof $p'$-core $O^p(G)\le O_{p'}(G)$. On the other hand the inclusion $O_{p'}(G)\le O^p(G)$ is banal because $O_{p'}(G)$ is a normal $p$'- subgroup of $G$.
So $O_{p'}(G)= O^p(G)$ and, being $\displaystyle\frac{G}{O^p(G)}$ a $p$-group, i have that $\displaystyle\frac{O_{p',p}(G)}{O_{p'}(G)}=O_p\biggl(\frac{G}{O_{p'}(G)}\biggr)=O_p\biggl(\frac{G}{O^p(G)}\biggr)=\frac{G}{O^p(G)}=\frac{G}{O_{p'}(G)}$ and in the end $G=O_{p',p}(G)$.
For curiosity: the $p',p$-core is generated by normal $p$-nilpotent subgroup of $G$ in analogy with Fitting supgroup.
I now ask help for $(\Leftarrow)$, please.
Ps. I thought I'd demonstrate that $O_{p'}(G)= O^p(G)$ and then obviously this subgroup is a normal $p$-complement.
If $O_{p',p}(G) = G$, then $O_p(G/O_{p'}(G)) = G/O_{p'}(G)$, so $G/O_{p'}(G)$ is a $p$-group, and hence $O_{p'}(G)$ is a normal $p$-complement in $G$.