Note: This is an alternative-proof question, since I know how to prove the result but I'm asking for a particular kind of proof. Why? For the fun of it!
Motivation:
I've been trying to give a reason behind the statement in this question from a particular perspective.
The Question:
Why, from a combinational group theoretical perspective on semidirect products, does the group $G$ given by
$$P=\langle r,s\mid r^8, s^2, srs=r^3\rangle$$
have two subgroups isomorphic to the Klein group?
Note: I separated $G$ from $P$ in my notation on purpose, just in case.
Thoughts:
By a standard result in combinatorial group theory, the group can be written as
$$\begin{align} G&\cong\langle r\mid r^8\rangle\rtimes_\varphi\langle s\mid s^2\rangle\\ &\cong \Bbb Z_8\rtimes_\varphi \Bbb Z_2, \end{align}$$
where $\varphi:\Bbb Z_2\to \operatorname{Aut}(\Bbb Z_8)$ is given by $\varphi(s)(r)=r^3$.
Are the two simply conjugate to each other?
According to GroupNames, there is at least one other way to write $G$ as a semidirect product. I would be happy using something other than the one above.
That's as far as I got.
I'm not entirely sure what's being asked, but you have a normal form $sr^k$ for elements in your group, via the last relation: \begin{equation*} r^ks = sr^{3k} \end{equation*} For such an element to have order $2$ you need \begin{equation*} 1=(sr^k)(sr^k) = (sr^ks)r^k = r^{4k} \end{equation*} So besides the central element $r^4$, there are four elements of order $2$: $\{sr^2, sr^4, sr^6, s\}$. That's enough for 2 Klein subgroups: \begin{align*} K_1 &= \{ 1, s, r^4, sr^4\}\\ K_2 &= \{ 1, sr^2, r^4, sr^6\} \end{align*} And these are conjugate: $r^{-1}K_1r=K_2$.