Neither direction of this is true in general. Can you write explicitly what the (presumably implicit) hypothesis is?

@verret I've edited my question.

So you are happy with $G/G'(p)\cong H/H'(p)$?

@DavidA.Craven Yes, as mentioned, I am only left to prove the equivalence.

$O^p(G)\leq G'(p)$.

@DavidA.Craven Yes. So if $G=O^p(G)$, then $G=G'(\pi)$ and so $H=H'(\pi)$. How to say $H=O^p(H)$ from here?

$O^p(H)\leq H'(p)$. I mean, that's just with a different letter.

@DavidA.Craven Yes, $O_p(H)\leq H'(p)=H$. Why does this imply $O^p(H)=H$?

@DavidA.Craven Can you give me a hint how to show $H'(p)\subseteq O^p(H)\cap H'$. Since $H'(p)$ is the smallest normal subgroup with Abelian $p$-factor group, I am trying to first show that $H/ (O^p(H)\cap H')$ is Abelian. This is equivalent to $H'\subseteq O^p(H)\cap H'$ or equivalently, $H'=O^p(H)$. I don't think this holds?

I'll correct it: $O^p(H)\cdot H′=H′(p)$.

@DavidA.Craven Why? And how does this give $O^p(H)=H$?

If you cannot see how $H'(p)<H$ iff both $O^p(H)=H$ and $H'<H$ then I cannot help you. And these are advanced group theory concepts. You thought about it for at most 5 minutes before asking why.

@DavidA.Craven Since $H'\subseteq O^p(H)H'$, so $H/O^p(H)H'$ is Abelian. Moreover, order of $H/O^p(H)H'$ divides order of $H/O^p(H)$, so $H/O^p(H)H'$ is a $p$-group. Finally, $O^p(H)H'$ is also a normal subgroup of $H$, so $H'(p)\subseteq O^p(H)H'$. This shows that $H=O^p(H)H'$. Correct?

Yes. So it's an immediate consequence of the statement about abelian $p$-group quotients.

@DavidA.Craven I still can't see why $H=O^p(H)$.

Right. $G$ has a $p$-quotient iff $G$ has an abelian $p$-quotient, because $p$-groups are soluble. We are assuming that $G$ has an abelian $p$-quotient iff $H$ has an abelian $p$-quotient. Finally, $H$ has an abelian $p$-quotient iff it has a $p$-quotient.

@DavidA.Craven Oh, this is the theorem that G is solvable iff G has a normal series all whose factors are Abelian iff G has a composition series all whose factors are prime order! I assumed that because Kurzweil and Stellmacher don't comment on the proof of the equivalence of $G=O^p(G)$ iff $H=O^p(H)$, it would be a trivial fact following from (immediate) previous assertions/definitions.

It is. $p$-groups always have abelian quotients.