I remember answering a very similar (possibly the same) question recently but I cannot find it.

Let $G$ be a central product of a cyclic group $M$ of order $4$ and the dihedral group $D_8$ of order $8$. So $|G|=16$. (In fact you get an isomorphic group if you replace $D_8$ by $Q_8$. A central product of $C_4$ with an extraspecial $2$-group is called a group of *symplectic type*.)

Then all subgroups of $G$ strictly between $M$ and $G$ have order $8$ and are isomorphic to $C_4 \times C_2$, and have the $C_2$ as the required subgroup $L$. But $M$ has no complement in $G$, so any subgroup $K$ with $K \cap M=1$ has order $1$ or $2$. (You need to assume that $K$ is nontrivial in the problem.) This example also satisfies your condition c), since the subgroups $D_8$ (and $Q_8$) are maximal and contain all minimal subgroups of $M$.

You can construct similar examples for any prime $p$ as a central product of $C_{p^2}$ with an extraspecial group of order $p^3$.