We have:
$MN//AB$
$\Rightarrow OM//AP$
We also have: $OP//AC$
$\Rightarrow OP//AM$
So: $AMOP$ is a parallelogram
Beside: $O$ is the intersection of the bisectors of $ΔABC$
$\Rightarrow OA$ is the bisector of $\widehat{ABC}$
$\Rightarrow OA$ is the bisector of $\widehat{MAP}$
Therefore: $AMOP$ is a rhombus
$\Rightarrow AM = OM = AP = OP$
By proving with the same way as above, we get:
$BNOQ$ is a rhombus
$\Rightarrow BN = NO = OQ = BQ$
Hence: $MN = OM + ON = AM + BN$
Moreover:
The perimeter of $ΔOPQ$ is:
$OP + PQ + OQ = AP + PQ + QB = AB\qquad (Q.E.D)$