Two different points $A$ and $B$  and a circle $\omega$ that passes through $A$  and $B$  are given. $P$  is a variable point on $\omega$  (different from $A$  and $B$). $M$ is a point such that $MP$  is the bisector of the angle $\angle{APB}$  ($M$ lies outside of $\omega$) and $MP=AP+BP$. Find the geometrical locus of $M$.
1. **Given Setup**: We are given two distinct points \( A \) and \( B \) and a circle \(\omega\) passing through \( A \) and \( B \). Let \( P \) be a variable point on \(\omega\) (different from \( A \) and \( B \)). \( M \) is a point such that \( MP \) is the bisector of the angle \(\angle APB\) and \( MP = AP + BP \).

2. **Ptolemy's Theorem**: According to Ptolemy's theorem, for a cyclic quadrilateral \( APBN \) where \( N \) is the midpoint of the arc \( \overarc{AB} \) not containing \( P \), we have:
   \[
   PA \cdot BN + PB \cdot AN = PN \cdot AB
   \]
   Since \( N \) is the midpoint of the arc \( \overarc{AB} \), \( AN = BN \). Therefore, the equation simplifies to:
   \[
   (PA + PB) \cdot AN = PN \cdot AB
   \]
   Dividing both sides by \( PN \cdot AN \), we get:
   \[
   \frac{PA + PB}{PN} = \frac{AB}{AN}
   \]
   This ratio is constant.

3. **Constructing Point \( A' \)**: Let \( A' \) be a point on the extension of the ray \( [NA \) such that \( AB = A'A \). Since \( A, N, B \) are fixed, \( A' \) is a constant point. We have:
   \[
   \frac{MP}{PN} = \frac{AA'}{AN} = \frac{AB}{AN}
   \]
   This implies that \( A'M \parallel AP \) and \(\angle A'MN = \angle APN\).

4. **Constructing Point \( B' \)**: Similarly, construct \( B' \) such that \( B' \) is a constant point. We have:
   \[
   \angle A'MB' = \angle APB = \text{Constant}
   \]
   Therefore, \( M \) lies on the arc \( \overarc{A'B'} \) of the circle passing through \( A', N, B' \) not containing \( N \).

5. **Second Midpoint \( N' \)**: There is another midpoint \( N' \) of the arc \( \overarc{AB} \) (the one containing \( P \)). Let the corresponding points be \( A'' \) and \( B'' \). By similar reasoning, \( M \) also lies on the arc \( \overarc{A''B''} \) of the circle passing through \( A'', N', B'' \) not containing \( N' \).

Conclusion:
The geometrical locus of \( M \) is the union of the two arcs \( \overarc{A'B'} \) and \( \overarc{A''B''} \) of the circles passing through \( A', N, B' \) and \( A'', N', B'' \) respectively, not containing \( N \) and \( N' \).

The final answer is the union of the arcs \( \boxed{ \overarc{A'B'} } \) and \( \overarc{A''B''} \).