Let $E\to M$ be an oriented rank 3 real vector bundle over a smooth 4-manifold $M$ and suppose $p_1(E)=0$. Why does this imply that the structure group of $E\to M$ can be reduced to a finite group $G$? I merely know that two $SO(3)$-bundles over a 4-manifold are isomorphic if and only if they have same $w_2$ and $p_1$.
By the way, my question is asserted in the last paragraph of p.353 of https://bpb-us-e2.wpmucdn.com/faculty.sites.uci.edu/dist/3/246/files/2011/03/23_PseudoFreeOrbifolds.pdf.
As Jonas pointed out in the comments, the conditions you list do not imply that $E$ admits a reduction of structure group to a finite group. However, the setting in the paper has an additional condition, namely that the bundle $E$ (explicitly $E_{\infty}$ in the paper) admits a self-dual connection $\nabla$ (explicitly $\nabla_{\infty}$ in the paper).
With respect to any connection $\nabla$, the first Pontryagin number of an orthogonal bundle $E$ over a closed oriented Riemannian four-manifold $(M, g)$ can be expressed using Chern-Weil theory by the formula
$$\langle p_1(E), [M]\rangle = \frac{1}{4\pi^2}\int_M\|R^{\nabla}_+\|^2 - \|R^{\nabla}_-\|^2d\mu_g$$
where $R^{\nabla}_{\pm}$ are the self-dual and anti-self-dual parts of $R^{\nabla}$, the curvature of $\nabla$. If $\nabla$ is a self-dual connection, then $R^{\nabla}_- = 0$, so $\langle p_1(E), [M]\rangle \geq 0$ with equality if and only if $\nabla$ is flat. In the case you're interested in, $\nabla$ is a self-dual connection and $p_1(E) = 0$, so $\nabla$ is flat.
A principal $G$-bundle with a flat connection admits a reduction of structure group to a discrete subgroup of $G$. Discrete subgroups of Lie groups are closed, see here, so if $G$ is compact, then the discrete subgroup is finite (a closed subset of a compact space is compact, and a compact discrete space is finite). In your case $G = SO(3)$, which is compact.
