proof
Let A,B,C,D have normals a,b,c,d at O, and let P=[p]
Since P lies on all the p-lines, p is perpendicular to a,b,c,d,
so the normals lie on the plane through O perpendicular to p.
It follows that a,b span this plane, so there are real numbers
α,β,γ,δ such that
(*)       c=αa+βb, d=γa+δb.
Note that axc,bxc,axd,bxd are multiples of p, so we can
discuss their ratios. If we multiply the equations (*) by c,d,
we see that βγ/αδ = {axc/bxc}/{axd/bxd}, and so is
independent of the choice of a,b,c,d.

Let L have normal u at O. Then A=[axp], etc.
If we multiply the equations (*) by [p], we see that
(A,B,C,D) = βγ/αδ, so is independent of the choice of L.