Take point $K$ on half of arc $AB$. We have that $KP=(AP+BP)\frac{AK}{AB}$. So we see that locus of the points M is the arc (if point P is only on the one arc).

why $KP=(AP+BP)\frac{AK}{AB}$

According to the Ptolemy's Theorem $KP.AB=AK.PB+KB.AP\Rightarrow KP=(AP+PB).\frac{AK}{AB}=MP.\frac{AK}{AB}$ ok. now, why locus of $M$ is an arc?

Let $B'$ and $A'$ be on $AX$ and $BX$ beyond $X$ such that $AB'=BA'=AB$ and $X$ is the middle of arc $AB$ without $P$ locus of $M$ is the arc $A'B'$ on circle $A'B'X$ without $X$. hint: i did it by using the centres $O_{1},O_{2}$ of the circles $PAB$ and $XA'B'$ and point $D$ which is the intersection of $PX$ and $AB$. It's not so hard it's just some ratio chase that uses similar triangles and $AX=BX$.

Not a different solution, but the complete result is a pair of arcs. We have $AP+BP$, angle bisector, a circumcircle. This is the special case of Ptolemy. Let $N$ be the midpoint of $\overarc {AB}$. By Ptolemy, $PA\cdot BN + PB\cdot AN = PN\cdot AB$ $\Rightarrow (PA+PB)\cdot AN = PN \cdot AB$ $\Rightarrow \frac{PA+PB}{PN} = \frac {AB}{AN} = \text{Constant.}$ Let $A'$ be a point on the extension of the ray $[NA$ such that $AB=A'A$. Since $A,N,B$ are constants, $A'$ is a constant point. We have $\frac{MP}{PN}=\frac{AA'}{AN} = \frac {AB}{AN}$ So $A'M \parallel AP$, and $\angle A'MN = \angle APN$. Similarly, construct $B'$ which is constant. We have $\angle A'MB' = \angle APB = \text{Constant}$. So $M$ is on the arc $%Error. "widearc" is a bad command. {A'B'}$. Since there is another $N$, call it $N'$, such that $N'$ is the midpoint of the arc $\overarc{AB}$. Let those points be $A''$ and $B''$. The complete solution is that the loci are the arc $A'B'$ of $(A'NB')$ not containing $N$ and the arc $A''B''$ of $(A''N'B'')$ not containing $N'$.

Why, $M$ lies outside of $ \omega $ ? There are $A,B$ and $P$ such that $PA+PB<2R$ and $ M(.)\in \omega $.