Start with equation for boost. Boost muon momentum and spin from rest frame.
Boosting in the 3 direction
\begin{equation}
A_{boost}=\begin{pmatrix}\gamma & 0 & 0 & -\gamma\beta_3\cr
                         0 & 1 & 0 & 0\cr
                         0 & 0 & 1 & 0\cr
                         -\gamma\beta_3& 0 & 0 & \gamma\end{pmatrix}
\end{equation}
The momentum four vector of the pion in its rest frame is
$$p = (m_\pi,0,0,0)$$
The sum of the muon 4 momentum $p_\mu$ and the neutrino four momentum $p_\nu$
$$p_\mu+p_\nu=p_\pi = E_\mu + E_\nu, \bfp_\mu+\bfp_\nu$$
$$\rightarrow \bfp_\mu=-\bfp_\nu$$
Since $E_\nu =|\bfp_\nu c|=|\bfp_\mu c|$ and $E_\mu=\sqrt{(m_\mu c^2)^2+(\bfp_\mu c)^2}$
then
$$|\bfp_\mu c|+\sqrt{(m_\mu c^2)^2 + (\bfp_\mu c)^2} = m_\pi c^2$$ 
It follows that 
$$(m_\mu c^2)^2 + (p_\mu c)^2 = (m_\pi c^2)^2 + (p_\mu c)^2 - 2 m_\pi p_\mu c^3$$
and
$$|\bfp_\mu| = \frac{(m^2_\pi-m^2_\mu)c}{2 m_\pi}$$
Also
\begin{eqnarray*}
E^2_\mu &=& \bfp_\mu^2 c^2 + m_\mu^2 c^4\\ 
&=& \frac{(m_\pi^2-m_\mu^2)^2 c^4}{4 m_\pi^2} +m_\mu^2 c^4\\
&=& \frac{1}{4}m_\pi^2 c^4 +\frac{1}{2} m_\mu^2 c^4 +\frac{m_\mu^4 c^2}{4m_\pi^2}\\
&=& \frac{(m_\pi^2 +m_\mu^2)^2c^4}{4 m_\pi^2}\\
\rightarrow E_\mu&=& \frac{(m_\pi^2+m_\mu^2)c^2}{2m_\pi}
\end{eqnarray*}
The direction of the momentum of the muon is not specified. If we suppose polar coordinates with
z in the direction of the boost then in the pion rest frame, 
$$p_\mu = \left[E_\mu,|\bfp_\mu|c(\cos\phi\sin\theta, \sin\phi\sin\theta, \cos\theta)\right]$$ 
and in the boosted frame.
$$p^\prime_\mu = \left[\gamma(E_\mu-\beta_3 |\bfp_\mu c|\cos\theta,
|\bfp_\mu c|(\cos\phi\sin\theta, \sin\phi\sin\theta), -\gamma(\beta_3 E_\mu-|\bfp_\mu c|\cos\theta)\right]$$ 
