Relativistic momentum in pion and muon

Using natural units $c=1$ what that relation implies is that the particles are on-shell. Whenever a particle is on-shell you have that, given the particle 4-momentum $p_mu$, $$p_mu p^mu equiv p^2 = m^2$$ where $m$ is the particle's mass.

If you put back the velocity of light, you'll get $$p_mu p^mu equiv p^2 = m^2c^4$$

An on-shell particle of mass $m$ and velocity ${bf v}$ has a four-velocity $u_mu = gamma(c,{bf v})$, with $gamma=(1-{bf v}^2/c^2)^{-1/2}$. Its relativistic momentum $p_mu$ is then given by the expression $p_mu = mu_mu$. The Lorentz invariant magnitude of $p_mu$ can then be calculated to be $$ p^2 equiv eta^{munu} p_mu p_nu = (p_0)^2 - (p_1)^2 - (p_2)^2 - (p_3)^2 = m^2c^2, $$ assuming the mostly-negative convention $eta^{munu}={rm diag}(1,-1,-1,-1)$.

$p^2$ means $p^mu p_mu=p_0^2-p_x^2-p_y^2-p_z^2$.

Usually, one denotes $p$ for the 4-vector, $vec{p}$ for the 3-vector and $p_i$ or $p^mu$ for the components.

So this is just the basic $p^2 = (mc)^2 = frac{E^2}{c^2}-vec{p}^2$ relation that one gets from $p^mu = (frac{E}{c}, vec{p})$ and basically defines mass.