5. $\triangle A B C$ is an equilateral triangle with side length 1, and $P A \perp$ plane $A B C$. Also, $P A=\frac{\sqrt{6}}{4}$. Let the point $A$'s reflection over plane $P B C$ be $A^{\prime}$, then the angle formed by line $A^{\prime} C$ and $A B$ is $\qquad$
5. $90^{\circ}$ As shown in the figure, let $A A^{\prime}$ intersect the plane $P B C$ at point $O$, and take the midpoint of $B C$ as $D$, then point $O$ lies on $P D$.

In the right triangle $\triangle P A D$, $A O \perp P D, P A=\frac{\sqrt{6}}{4}, A D=\frac{\sqrt{3}}{2}$,
then $P D=\sqrt{P A^{2}+A D^{2}}=\frac{3 \sqrt{2}}{4}$.
By the area relationship, we get $A O=\frac{P A \cdot A D}{P D}=\frac{1}{2}$, hence $A^{\prime} A=2 A O=1$,
Since point $A^{\prime}$ is symmetric to $A$ with respect to the plane $P B C$,
we have $A^{\prime} B=A B=1, A^{\prime} C=A C=1$,
Therefore, the tetrahedron $A^{\prime} A B C$ is a regular tetrahedron.
Thus, the angle formed by the skew lines $A^{\prime} C$ and $A B$ is $90^{\circ}$.