$$
\begin{array}{l}
\text { 3. Given } E F=C E \\
=C F, E A=B F \\
=2 A B, A B=B D \\
=D A, \text { and } A P= \\
C P=B Q=C Q \\
=P D=D Q=1 \text {. Then the line segment } B D=\text {. }
\end{array}
$$
From the problem, we know that the conditions determine the symmetry of the entire figure.

As shown in Figure 4, connect $C A, C B, C D$.
It is easy to prove
$$
\begin{array}{l}
\triangle A E C \cong \\
\triangle B F C \\
\Rightarrow C A=C B .
\end{array}
$$

Since $D A=D B, C D=C D$, therefore, $\triangle A D C \cong \triangle B D C$.
Thus, $\angle C D A=\angle C D B$
$$
=\frac{1}{2}\left(360^{\circ}-\angle A D B\right)=150^{\circ} \text {. }
$$

Given $A P=C P=P D$, by Theorem 1, $P$ is the circumcenter of $\triangle A C D$.
Then, by Property 2(2),
$$
\angle C D A=180^{\circ}-\frac{1}{2} \angle A P C \text {. }
$$

Thus, $\angle A P C=60^{\circ}$.
Since $A P=C P$, $\triangle A P C$ is an equilateral triangle.
Therefore, $C A=A P=1$.
Extend $C D$ to intersect $A B$ at point $O$. It is easy to know that $C O \perp A B, A O=B O=\frac{1}{2} A B$,
$$
C O=\frac{\sqrt{3}}{2} E F=\frac{5 \sqrt{3}}{2} A B \text {. }
$$

In the right triangle $\triangle A O C$, by the Pythagorean theorem, $A O^{2}+C O^{2}=C A^{2}$,
which is $\left(\frac{1}{2} A B\right)^{2}+\left(\frac{5 \sqrt{3}}{2} A B\right)^{2}=1$.
Solving this, we get $A B=\frac{\sqrt{19}}{19}$.
Therefore, $B D=\frac{\sqrt{19}}{19}$.