The only way π− is emitted at rest in the lab system is when it is emitted at θ1∗ = 180◦ in the CMS (rest frame of K◦) with with the same speed as K◦ in the lab system. In that case π− will be emitted at θ∗2 = 0◦ in the CMS.
The energy released
Q = 498 − 2 × 140 = 218MeV
As the product particles are identical, each pion carries half of the enrgy, 109 MeV