Draw the diameter $HK$ that passes through $P$ and let $C$ and $D$ be the mid-points of $PK$ and $PH$, respectively.

Since $E$ is the mid-point of $AP$, $EC$ is parallel to $AK$ and $ED$ is parallel to $AH$. This implies that $\angle DEC$ is similar to $\angle HAK$, and so $\angle DEC = \angle HAK$. Because $HK$ is a diameter, this shows that $\angle DEC = 90^\circ$. This implies that $E$ is on the circle with diameter $CD$ the radius of which is equal to $|CD|/2$. Because
$$|CD| = |CP| + |PD| = \frac{|KP|}{2} + \frac{|PH|}{2} = \frac{|HK|}{2} = r$$
hence, $E$ is on the circle with radius $r/2$ which has $CD$ as a diameter.

Let $Q$ be the mid-point of $PO$. Then
$$\frac{|PO|}{|PQ|} = \frac{2}{1} = \frac{|PA|}{|PE|},$$
hence, $OA$ is parallel to $QE$. This implies that $\frac{|OA|}{|QE|} = \frac{1}{2}$, i.e. $|QE| = |OA|/2 = r/2$. This means that all possible $E$ are on the circle centre $Q$

of radius $r/2$. This circle is the image of the original circle under the homothety with centre $P$ and factor $1/2$.