Yes, a $T_3$-space is paracompact iff every open cover of it has a $\sigma$-discrete open refinement. In the course of this answer I proved that if $X$ is a paracompact Hausdorff space, then every open cover of $X$ has an open barycentric refinement, and in the course of this answer I proved that a barycentric open refinement of a barycentric open refinement of an open cover is an open star refinement of the original open cover. Thus, it suffices to show that if every open cover of $X$ has an open star refinement, then every open cover of $X$ has a $\sigma$-discrete open refinement. This is Lemma $\mathbf{5.1.16}$ in R. Engelking, General Topology.
Assume that every open cover of $X$ has an open star refinement, and let $\mathscr{U}=\{U_\xi:\xi<\kappa\}$ be an open cover of $X$. Let $\mathscr{U}_0=\mathscr{U}$, and for $n\in\Bbb N$ let $\mathscr{U}_{n+1}$ be an open star refinement of $\mathscr{U}_n$. For $\xi<\kappa$ and $n\in\Bbb Z^+$ let
$$U_\xi(n)=\{x\in X:\operatorname{st}(V,\mathscr{U}_n)\subseteq U_\xi\text{ for some open nbhd }V\text{ of }x\}\;.$$
Then for each $n\in\Bbb Z^+$, $\{U_\xi(n):\xi<\kappa\}$ is an open refinement of $\mathscr{U}$ covering $X$.
Suppose that $n\in\Bbb Z^+$, $\xi<\kappa$, and $x\in U_\xi(n)$. Suppose further that $x\in U\in\mathscr{U}_{n+1}$. $\mathscr{U}_{n+1}$ star refines $\mathscr{U}_n$, so there is a $W\in\mathscr{U}_n$ such that $\operatorname{st}(U,\mathscr{U}_{n+1})\subseteq W\subseteq\operatorname{st}(x,\mathscr{U}_n)\subseteq U_\xi$, and by definition $U\subseteq U_\xi(n+1)$. Thus, $\operatorname{st}(x,\mathscr{U}_{n+1})\subseteq U_\xi(n+1)$.
Now for each $\eta<\kappa$ and $n\in\Bbb Z^+$ let
$$V_\eta(n)=U_\eta(n)\setminus\operatorname{cl}\bigcup_{\xi<\eta}U_\xi(n+1)\;,$$
and let $\mathscr{V}_n=\{V_\xi(n):\xi<\kappa\}$; clearly each $\mathscr{V}_n$ is an open refinement of $\mathscr{U}$. I claim that each $\mathscr{V}_n$ is a discrete collection, and that $\mathscr{V}=\bigcup_{n\in\Bbb Z^+}\mathscr{V}_n$ covers $X$.
Fix $n\in\Bbb Z^+$, and suppose that $\xi<\eta<\kappa$; then $V_\eta(n)\subseteq X\setminus U_\xi(n+1)$, and $V_\xi(n)\subseteq U_\xi(n)$. Suppose that $x\in V_\xi(n)$ and $y\in V_\eta(n)$. Then $\operatorname{st}(x,\mathscr{U}_{n+1})\subseteq U_\xi(n+1)$, while $y\notin U_\xi(n+1)$, so no member of $\mathscr{U}_{n+1}$ contains both $x$ and $y$, and it follows that $\mathscr{V}_n$ is discrete.
Let $x\in X$. Let $\eta=\min\{\xi<\kappa:x\in U_\xi(n)\text{ for some }n\in\Bbb Z^+\}$; this exists because each $\{U_\xi(n):\xi<\kappa\}$ for $n\in\Bbb Z^+$ is an open cover of $X$. Fix $n\in\Bbb Z^+$ such that $x\in U_\eta(n)$. If $\xi<\eta$, then $x\notin U_\xi(n+2)$. Suppose that $y\in\bigcup_{\xi<\eta}U_\xi(n+1)$; then $\operatorname{st}(y,\mathscr{U}_{n+2})\subseteq U_\xi(n+2)$, so $x\notin\operatorname{st}(y,\mathscr{U}_{n+2})$, and hence $y\notin\operatorname{st}(x,\mathscr{U}_{n+2})$. It follows that $\operatorname{st}(x,\mathscr{U}_{n+2})$ is an open nbhd of $x$ disjoint from $\bigcup_{\xi<\eta}U_\xi(n+1)$ and hence that $$x\in U_\eta(n)\setminus\operatorname{cl}\bigcup_{\xi<\eta}U_\xi(n+1)=V_\eta(n)\;.$$ Thus, $\mathscr{V}$ covers $X$.
