I think it is supposed to be $AM=MP+PB \ \ (1)$ and $BM=MP+PA \ \ (2)$ $AM=MP+PB \Rightarrow $ $AP=AM+MP=2MP+PB \Rightarrow $ $AP>PB$ Similarly, from $(2)$ we get $AP<PB$. Let $CD$ be the perpendicular diameter to $AB$. Suppose that $P$ is such that $AP>PB$ so we use the formula $(1)\ \ AM=MP+PB$ On the extension of $AP$ we get the point $E$ such that $PE=PB$. Then $AM = MP+PE \Rightarrow AM = ME\Rightarrow M$ is the midpoint of $AE$ Now suppose that $P\in\stackrel{\frown}{BD}$. We'll prove that the point $D$ is on the perpendicular bisector of $BE$. Indeed, the external angle bisector of the isosceles triangle $PBE$ is the line $PC$. But $PC\perp PD$, so the internal angle bisector of $\triangle PBE$ is the line $PD$. But $PBE$ is isosceles, so $PD$ is also the perpendicular bisector of its base $BE$. On the other hand, $CD$ is the perpendicular bisector of $AB$, so we get $DA=DB=DE$. So the point $E$ is moving on the circle $(D,DA)$ As a result, $DM\perp AM$, so $M$ lies on the circle with diameter $AD$ Similarly: $P\in \stackrel{\frown}{BC}\Rightarrow M$ on the circle with diameter $AC$. $P\in \stackrel{\frown}{AC}\Rightarrow M$ on the circle with diameter $BC$. $P\in \stackrel{\frown}{AD}\Rightarrow M$ on the circle with diameter $BD$. The locus consists of the red arcs.

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