Answer
$Q= -450 \;\rm MeV$
$KE_{\pi^-}= 767.808 \;\rm MeV$
Work Step by Step
The $Q$-value of the given reaction is given by
$$Q=Q_{\pi^-}+Q_{\rm p}-Q_{\Lambda^0}-Q_{\rm K^0}$$
Thus,
$$Q=(m_{\pi^-}+m_{\rm p}-m_{\Lambda^0}-m_{\rm K^0})c^2$$
Plugging the known (see table 32-2);
$$Q= 135+ 938.3-1115.6-497.7 =\bf \color{red}{\bf -450}\;\rm MeV$$
Since $\pi^-$ hits the proton when it was at rest, so the initial kinetic energy of the proton is zero.
And according to the conservation of energy;
$$E_{\pi^-}+E_{\rm p}=E_{\Lambda^0}+E_{\rm K^0}$$
where $E_{\rm p}=\overbrace{KE_{\rm p}}^{0}+m_{\rm p}c^2$
$$E_{\pi^-}+m_{\rm p}c^2=E_{\Lambda^0}+E_{\rm K^0}\tag 1$$
Since the initial momentum is not zero, and since the two products are having the same speed, then the two products are moving in the same direction as one particle because the sum of their momentum must be equal to the initial momentum of $\pi^-$.
Thus; $$m_{\rm proudcuts}=m_{\Lambda^0}+m_{\rm K^0}\tag 2$$
and hence, we can rewrite (1) as follows;
$$E_{\pi^-}+m_{\rm p}c^2=E_{\rm products} \tag 3$$
Acording to the conservation of momentum;
$$p_{\pi^-}=p_{\Lambda^0}+p_{\rm K^0}=p_{\rm proudcuts}$$
Multiplying both sides by $c$ and then squaring both sides.
$$p^2_{\pi^-}c^2= p^2_{\rm proudcuts}c^2$$
where $p^2c^2=E^2-m^2c^4$. So that;
$$E_{\pi^-}^2-m_{\pi^-}^2c^4= E_{\rm proudcuts}^2-m_{\rm proudcuts}^2c^4$$
Plugging from (3);
$$E_{\pi^-}^2-m_{\pi^-}^2c^4= (E_{\pi^-}+m_{\rm p}c^2)^2-m_{\rm proudcuts}^2c^4$$
$$\color{red}{\bf\not}E_{\pi^-}^2-m_{\pi^-}^2c^4= \color{red}{\bf\not}E_{\pi^-}^2+2 m_{\rm p}c^2E_{\pi^-}+m^2_{\rm p}c^4 -m_{\rm proudcuts}^2c^4$$
Solving for $E_{\pi^-}$;
$$ m_{\rm proudcuts}^2c^4 -m_{\pi^-}^2c^4-m^2_{\rm p}c^4 = 2 m_{\rm p}c^2E_{\pi^-} $$
$$ E_{\pi^-}=\dfrac{m_{\rm proudcuts}^2c^4 -m_{\pi^-}^2c^4-m^2_{\rm p}c^4 } { 2 m_{\rm p}c^2} $$
Recalling that $E_{\pi^-}=KE_{\pi^-}+m_{\pi^-}c^2 $, so that
$$ KE_{\pi^-}+m_{\pi^-}c^2=\dfrac{m_{\rm proudcuts}^2c^4 -m_{\pi^-}^2c^4-m^2_{\rm p}c^4 } { 2 m_{\rm p}c^2} $$
Thus;
$$ KE_{\pi^-}=\dfrac{m_{\rm proudcuts}^2c^4 -m_{\pi^-}^2c^4-m^2_{\rm p}c^4 } { 2 m_{\rm p}c^2} -m_{\pi^-}c^2$$
Plugging the known;
$$ KE_{\pi^-}=\dfrac{(1115.6+497.7)^2 -139.6^2 - 938.3^2 } { 2\times 938.3} - 139.6$$
$$ KE_{\pi^-}=\color{red}{\bf 767.808}\;\rm MeV$$
