If $G$ is a group with order $364$, then it has a normal subgroup of order $13$

Suppose $G$ is a group of order $364 = 2^2 \cdot 7 \cdot 13$.

As you noted, there is a normal subgroup $P$ of order $7$ by Sylow's theorems. By Cauchy's theorem, there is a subgroup $K$ of order $13$ in $G$.

Now $P$ is normal, so $PK$ is a subgroup of order $7 \cdot 13$. Since $PK$ has index $4$, there exists a homomorphism $\phi: G \rightarrow S_4$ with $\operatorname{Ker}(\phi)$ contained in $PK$. You can see that this implies that $\operatorname{Ker}(\phi)$ equals $PK$ and thus $PK$ is a normal subgroup. Since every group of order $7 \cdot 13$ is cyclic and subgroups of a normal cyclic subgroup $PK$ are also normal in $G$, the subgroup $K$ is a normal subgroup.

After m.k.'s proof another one. There is indeed exactly one $7$-sylow subgroup, let's call is $S$: it is normal in $G$. There is also exactly one Sylow $13$-group in $G/S$ because the number divides $4$ and has residue $1$ modulo $13$. Take some generator of this unique Sylow $13$-subgroup of $G/S$. It has exactly $7$ antecedents in $G$. Also, any Sylow $13$-subgroup of $G$ intersects $S$ trivially so projects isomorphically onto the unique Sylow Subgroup of $G/S$. Each Sylow $13$-subgroup of $G$ thus contains one of the seven antecedents mentioned above, and two distinct Sylow $13$-subgroups of $G$ must intersect trivially, thus there areat most $7$ Sylow $13$-subgroups in $G$. But there can only be $1$ or $14$ so there is exactly one and it is normal.