Answer
$$ KE_{\mu} = \dfrac{ (m_{\pi} -m_{\mu})^2c^2}{2m_{\pi} }$$
See the detaied answer below.
Work Step by Step
Since the pi meson decays at rest to a muon and a neutrino (with negligible mass), so the net momentum must be zero and hence the neutrino and the muon must have the same momentum magnitude but in opposite directions.
$$ p_{\mu}=p_{\nu}=p\tag 1$$
The net energy is assumed to be constant since we assume that the system is isolated.
Thus,
$$E_{\pi}=E_{\mu}+E_{\nu}\tag 2$$
Where the energy of the neutrnio is given by
$$E_{\nu}=\sqrt{p_{\nu}^2c^2+m_{\nu}^2c^4}$$
Recall that the neutrino has a negligible mass, so $m_{\nu}=0$
$$E_{\nu}=\sqrt{p_{\nu}^2c^2+(0)c^4}$$
$$E_{\nu}=p_{\nu}c\tag 3$$
The energy of the muon is given by
$$E_{\mu}=\sqrt{p_{\mu}^2c^2+m_{\mu}^2c^4}\tag 4$$
And the energy of the pi meson is given by
$$E_{\pi}=\sqrt{p_{\pi}^2c^2+m_{\pi}^2c^4} $$
and since it was at rest, so $p_{\pi}=0$;
$$E_{\pi}=\sqrt{(0)^2c^2+m_{\pi}^2c^4} $$
$$E_{\pi}=m_{\pi}c^2\tag 5 $$
Now we need to plug (3), (4), and (5) into (2);
$$m_{\pi}c^2=\sqrt{p_{\mu}^2c^2+m_{\mu}^2c^4}+p_{\nu}c $$
and from (1);
$$m_{\pi}c^2=\sqrt{p^2c^2+m_{\mu}^2c^4}+pc $$
$$m_{\pi}c^2-pc =\sqrt{p^2c^2+m_{\mu}^2c^4}$$
Squaring both sides;
$$\left[m_{\pi}c^2-pc \right]^2= p^2c^2+m_{\mu}^2c^4 $$
$$ m_{\pi}^2c^4- 2m_{\pi}c^3p +\color{blue}{p^2c^2} =\color{blue}{ p^2c^2}+m_{\mu}^2c^4 $$
$$ m_{\pi}^2c^4- 2m_{\pi}c^3p =m_{\mu}^2c^4 $$
Solving for $p$;
$$ p =\dfrac{m_{\pi}^2c^4-m_{\mu}^2c^4}{2m_{\pi} c^3} $$
Notice the common factor $c^2$;
$$ p =\dfrac{m_{\pi}^2c^2-m_{\mu}^2c^2}{2m_{\pi} c } \tag 6$$
We know that the kinetic energy of the muon is given by
$$E_{\mu}=KE_{\mu}+m_{\mu}c^2$$
Thus,
$$KE_{\mu} =E_{\mu}-m_{\mu}c^2\tag 7$$
And from (2);
$$ E_{\mu} =E_{\pi}-E_{\nu} $$
Plugging from (3) and (5);
$$ E_{\mu} =m_{\pi}c^2 -pc $$
Plugging into (7);
$$KE_{\mu} =m_{\pi}c^2 -pc -m_{\mu}c^2=(m_{\pi}c^2 -m_{\mu}c^2)-pc $$
Plugging $p$ from (6);
$$KE_{\mu} =(m_{\pi}c^2 -m_{\mu}c^2)- \dfrac{m_{\pi}^2c^2-m_{\mu}^2c^2}{2m_{\pi} }$$
$$KE_{\mu} = \dfrac{2m_{\pi}^2c^2- 2m_{\pi}m_{\mu}c^2- m_{\pi}^2c^2-m_{\mu}^2c^2}{2m_{\pi} }$$
$$KE_{\mu} = \dfrac{ m_{\pi}^2c^2- 2m_{\pi}m_{\mu}c^2 -m_{\mu}^2c^2}{2m_{\pi} }$$
$$KE_{\mu} = \dfrac{ [m_{\pi}^2 - 2m_{\pi}m_{\mu} -m_{\mu}^2]c^2}{2m_{\pi} }$$
$$\boxed{KE_{\mu} = \dfrac{ (m_{\pi} -m_{\mu})^2c^2}{2m_{\pi} }}$$
