I'm curious to know if, given four points $a, b, c, d$, you can always find a point $p$ such that last lines $pa, pb, pc, pd$ form equal angles pairwise.
I'd also appreciate resources on 3d geometry especially if there is an analogue of inscribed angles for a circle.
No, there is not. Take $ABC$ and $BCD$ as two equilateral triangles, let $M$ be the midpoint of $BC$, $\pi_A,\pi_D,\pi_M$ be the planes where $ABC,BCD$ and $ADM$ lie. The locus $l_A$ of points $P$ such that $P$ "see" all the sides of $ABC$ under the same angle is a line through the center of $ABC$ orthogonal to $\pi_A$. By symmetry, $l_A$ and $l_D$ meet on $\pi_M$, above $AD$, while the loci associated with the $ABD$ and $ACD$ faces meet on $\pi_M$, too, but below $AD$, so there is no 3d-equivalent of the Steiner-Torricelli-Fermat point in this configuration. This is even more evident if you take $ABCD$ as a degenerate tetrahedron, in which $A,B,C,D$ all lie on the same plane and $ABC,BCD$ are equilateral triangles, with $A$ and $D$ symmetric with respect to the midpoint of $BC$.