\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\begin{document}
\section{3D Projection}
\def\minY{-2} \def\maxY{2}
\def\minZ{-2} \def\maxZ{6}
\def\screenZ{2}
\def\eyeZ{0} \def\eyeY{0}
\def\pZ{3} \def\pY{1}
\begin{tikzpicture}
  \draw[-stealth] (\minZ,0) -- (\maxZ,0) node[above]{$z$};
  \draw[-stealth] (0,\minY) -- (0,\maxY) node[above]{$x$};
  \draw[fill] (\eyeZ, \eyeY) circle (.5ex) node[above]{eye $E$};
  \draw[dotted,thick] (\screenZ, \minY) -- (\screenZ, \maxY) node[above]{screen};
  \draw[dotted] (\eyeZ, \eyeY) -- (\pZ, \pY);
  \draw[dotted] (\eyeZ, \eyeY) -- (\pZ, -\pY);
  \draw[fill] (\pZ, \pY) circle (.5ex) node[above]{$p_1$};
  \draw[fill] (\pZ, -\pY) circle (.5ex) node[above]{$p_2$};
  \draw[fill] (\screenZ, {(\pY - \eyeY)/(\pZ - \eyeZ)*(\screenZ - \eyeZ) + \eyeY}) circle (.5ex) node[above]{$p_1'$};
  \draw[fill] (\screenZ, {-(\pY - \eyeY)/(\pZ - \eyeZ)*(\screenZ - \eyeZ) + \eyeY}) circle (.5ex) node[above]{$p_2'$};
\end{tikzpicture}

When we refer to $p$ we mean either $p_1$ or $p_2$. When we refer to $p'$ we mean either $p'_1$ or $p'_2$.

\begin{align}
  & p = (p_x, p_y, p_z) \\
  & p'= (p'_x, p'_y, S_z) \\
  & E = (E_x, E_y, E_z) \\
\end{align}

The formula to find $p'$

\begin{align}
  & p_y' = \frac{(p_y - E_y)(S_z - E_z)}{(p_z - E_z)} + E_y \\
  & p_x' = \frac{(p_x - E_x)(S_z - E_z)}{(p_z - E_z)} + E_x \\
\end{align}

If we assume that $E = (0, 0, 0)$ and $S_z = 1$

\begin{align}
  & p_y' = \frac{p_y}{p_z} \\
  & p_x' = \frac{p_x}{p_z} \\
\end{align}

\end{document}