I think there is no such map. We can assume that it is smooth. Consider the incidence variety $Z=\{(p,\ell)\in \mathbb{P}^3\times \mathbb{G}(2,4)\,|\, x\in\ell\} $. The projection $p:Z\rightarrow \mathbb{P}^{3}$ is a $\mathbb{P}^2$-bundle, in fact it is the projective tangent bundle to $\mathbb{P}^3$. You are asking for a section of this bundle; that would give a (complex) line bundle $M$ on $\mathbb{P}^3$ which is a direct summand of the tangent bundle $T_{\mathbb{P}^3}$. This implies $c_3(T_{\mathbb{P}^3}\otimes M^{-1})=0$; but line bundles on $\mathbb{P}^3$ are powers of the tautological line bundle $L$, and an easy calculation gives $c_3(T_{\mathbb{P}^3}\otimes L^k)=[(k+1)^4-1]/k $, which is always $\neq 0$.

I think there is no holomorphic such map. Consider the incidence variety $Z=\{(p,\ell)\in \mathbb{P}^3\times \mathbb{G}(2,4)\,|\, x\in\ell\} $. The projection $p:Z\rightarrow \mathbb{P}^{3}$ is a $\mathbb{P}^2$-bundle, in fact it is the projective tangent bundle to $\mathbb{P}^3$. You are asking for a section of this bundle; that would give a line bundle $M$ on $\mathbb{P}^3$ which is a subbundle of the tangent bundle $T_{\mathbb{P}^3}$. Computing $c_3$ one sees that this line bundle must be $\mathcal{O}_{\mathbb{P}^3}(2)$; but $H^0(T_{\mathbb{P}^3}(-2))$ is zero, so $\mathcal{O}_{\mathbb{P}^3}(2)$ does not inject into $T_{\mathbb{P}^3}$.

I do not know if there exists a continuous section (contrary to what I wrote before editing).