If $|x - 2| = p$, where $x < 2$, then $x - p =$
$\textbf{(A)} \ -2 \qquad \textbf{(B)} \ 2 \qquad \textbf{(C)} \ 2-2p \qquad \textbf{(D)} \ 2p-2 \qquad \textbf{(E)} \ |2p-2|$
When $x < 2,$ $x-2$ is negative so $|x - 2| = 2-x = p$ and $x = 2-p$.
Thus $x-p = (2-p)-p = 2-2p$.
$\boxed{\mathbf{(C)}\ \ensuremath{2-2p}}$