Let $ T/N = \Phi(G/N) $. Why $ K \cap T = 1 $? Let $ G $ be a soluble group whit $ \Phi(G) = 1 $ that $ \Phi(G) $ is a Frattini subgroup of $ G $. Let $ K $ and $ N $ are minimal normal subgroup of $ G $ that $ K \neq N $. Let $ T/N = \Phi(G/N) $. Why $ K \cap T = 1 $? 
 A: Your statement, after little thinking, found interesting. It is possibly equivalent to say that with your hypothesis, there is a maximal subgroup of $G$ which contains one of $K$ and $N$ but not the other. 
I tried to prove this in following way: we use following theorem from Huppert-Endliche Gruppen Chapter III, Theorem 4.4.

Let $G$ be a solvable group and $A$ be an abelian normal subgroup such that $A\cap \Phi(G)=1$. Then $A$ has a complement in $G$ (i.e. $\exists$ $H\leq G$ such that $AH=G$ and $A\cap H=1$.)

(1) In a finite group, minimal normal subgroup is direct product of simple subgroups (isomorphic within themselves).
(2) Since $G$ is solvable, $K$ would be solvable and hence by (1), $K$ is direct product of cyclic (sub)groups of same prime order. Same is true for $N$. (so $K,N$ are abelian also.)
(3) Since $K$ and $N$ are distinct minimal normal subgroups, we have $K\cap N=1$, hence every element of $K$ commutes with every element of $N$, and so $KN$ is an abelian normal subgroup.
(4) Since $KN$ is abelian and $KN\cap \Phi(G)=1$, there is a complement $H$ of $KN$ in $G$.
(5) Thus $G=(KN).H$ with $KN\cap H=1$. Hence $N\cap H=1$, and as $N\trianglelefteq G$, we have $NH$ is a subgroup of $G$.  In particular, $|G|=|KN|.|H|=|K|.|N|.|H|$.
(6) $NH$ is proper subgroup since $|NH|=|N|.|H|<|G|$. Let $M$ be any (proper) maximal subgroup of $G$ which contains $NH$. 
(7) Then $M$ can not contain $K$ (otherwise, $M$ will contain $K.N.H=G$). 
In particular, $M$ is a maximal subgroup of $G$ such that 


*

*$M$ contains $N$;

*$M$ do not contain $K$.
(8) Now coming to your normal subgroup $T$. This $T$ is intersection of those maximal subgroups of $G$ which contain $N$.
(9) If $K\cap T\neq 1$ then by minimality of $K$ we must have $K\subseteq T$, i.e. whenever a maximal subgroup contains $N$, it contains $K$ as well. This is contradiction by (7).
