5. In square $ABCD$, $M$ is the midpoint of $AD$, and $N$ is the midpoint of $AB$. Fold $\triangle CDM$ and $\triangle CBN$ along $CM$ and $CN$ respectively, so that $CB$ and $CD$ coincide. Let point $B$ coincide with point $D$ at point $P$, and the midpoint of $DM$ becomes the midpoint of $PM$, denoted as $T$. Then the angle formed by the skew lines $CT$ and $PN$ is ( ).
(A) $45^{\circ}$
(B) $60^{\circ}$
(C) $75^{\circ}$
(D) $90^{\circ}$
$5 .(\mathrm{D})$.
In square $ABCD$, connect $AC$ and $MN$, and let their intersection be $S$. Since $M$ is the midpoint of $AD$ and $N$ is the midpoint of $AB$, by symmetry, $S$ is the midpoint of $MN$. Connect $TS$, from $\triangle MPN$ we get $TS \parallel PN$. Therefore, $\angle CTS$ is the angle formed by the skew lines $CT$ and $PN$.
Taking the side length of the square as 4 units, then
$$
\begin{array}{l}
CT=\sqrt{1^{2}+4^{2}}=\sqrt{17}, \\
TS=\frac{1}{2} PN=\frac{1}{2} BN=1, \\
CS=CA-SA=4 \sqrt{2}-\sqrt{2}=3 \sqrt{2} .
\end{array}
$$

Thus, we get $CT^{2}+TS^{2}=CS^{2}$.
$$
\therefore \angle CTS=90^{\circ} \text {. }
$$

That is, the angle formed by the skew lines $CT$ and $PN$ is $90^{\circ}$.