3. In trapezoid $ABCD$, $BC \parallel AD$, $AB=AD$, $\angle ABC=\frac{2\pi}{3}$, $\angle BCD=\frac{\pi}{2}$. $\triangle ABD$ is folded along $BD$, and the projection of point $A$ on plane $BCD$ is point $P$. It is known that the cosine of the angle between $AB$ and $CD$ is $\frac{\sqrt{3}}{6}$. Then the cosine of the angle between $BP$ and $CD$ is
3. $\frac{1}{2}$.

Let the angle between $AB$ and plane $BCD$ be $\theta$, and the angle between $BP$ and $BD$ be $\alpha$.
By the triple cosine theorem, we have
$$
\begin{array}{l}
\cos \frac{\pi}{3}=\cos \theta \cdot \cos \alpha, \\
\frac{\sqrt{3}}{6}=\cos \theta \cdot \cos \left(\frac{\pi}{6}+\alpha\right) .
\end{array}
$$

Thus, $\alpha=\frac{\pi}{6}$.
Therefore, the angle between $BP$ and $CD$ is $\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}$, and its cosine value is $\frac{1}{2}$.