4.17.

As shown in Figure 4, let the projections of points $A$ and $D$ on plane $\alpha$ be $O$ and $C$, respectively, then we have $A O = C D = 8$.
$$\begin{array}{l} \because A B = 10, \\ \therefore O B = 6 \text{ (constant). } \end{array}$$

Therefore, the trajectory of point $B$ in $\alpha$ is a circle with $O$ as the center and 6 as the radius.
$$\begin{array}{l} \because C D \perp \alpha, \text{ then } \\ B D^{2} = B C^{2} + C D^{2} \\ = B C^{2} + 8^{2}, \end{array}$$
$\therefore$ The minimum value of $B D$ is to find the minimum value of $B C$.
Since point $C$ is outside the circle $\odot O$, connect $O C$ intersecting $\odot O$ at $B_{0}$. By plane geometry knowledge, we easily get
$$\begin{array}{l} (B C)_{\min} = B_{0} C = C O - O B_{0} = 21 - 6 = 15. \\ \text{ Hence } (B D)_{\min} = \sqrt{(B C)_{\min}^{2} + 8^{2}} = 17. \end{array}$$