Answer
The decay scheme is $p\longrightarrow \hspace{2mm}e^+ \hspace{2mm}+ \hspace{2mm}\nu_e$
where $p$ = proton
$e^+$ = positron
$\nu_e$ = electron-neutrino
This does not violate linear momentum conservation.
Work Step by Step
It would be sufficient to verify the linear momentum conservation law in the rest frame of the proton. In the following diagram, we have a proton at rest and it has initial linear momentum = $\vec{0}$
It decays to $e^+$ with linear momentum $\vec{m}$ and $\nu_e$ with linear momentum $-\vec{m}$. Thus, the total final linear momentum is $\vec{m}+(-\vec{m})=\vec{0}$
Thus, the decay scheme does not violate linear momentum conservation.
