= Solution
Restrict the quotient projection $\pi:G\to G/N$ to $H$. This is a <group homomorphism> with
$$
\ker(\pi|_H)=H\cap N.
$$
Its image is a nontrivial <subgroup> of the prime-order <quotient group>, because $H$ is not contained in $N$. By <Lagrange's theorem for finite groups>, the image has order $p$ and is all of $G/N$. The <first isomorphism theorem for groups> therefore gives
$$
\boxed{H\cap N\trianglelefteq H,\qquad [H:H\cap N]=p.}
$$
For the second assertion, partition $C$ into <conjugacy classes> for $N$. Because $N$ is normal, <conjugation> by $G$ permutes these smaller classes: $g\operatorname{Cl}_N(x)g^{-1}=\operatorname{Cl}_N(gxg^{-1})$. This <group action> is transitive since any two elements of $C$ are conjugate in $G$. Elements of $N$ fix every $N$-class, so the action factors through $G/N$.
If $m$ is the number of smaller classes, the <orbit-stabilizer theorem> for this transitive action gives $m\mid |G/N|=p$. Thus
$$
\boxed{C\text{ is one }N\text{-class, or a disjoint union of }p\text{ such classes}.}
$$
Distinct <conjugacy classes> are disjoint because they are <orbits of a group action>. This proves <conjugacy class splitting in a prime-index normal subgroup> without assuming all $G$-conjugations are already conjugations by $N$.
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