3. As shown in Figure 3, in the Cartesian coordinate system $x O y$, $A B$ is the diameter of the semicircle $\odot O$: $x^{2}+y^{2}=1(y \geqslant 0)$, $C$ is any point on the semicircle $\odot O$, and $A C$ is extended to point $P$ such that $C P=C B$. When point $C$ moves from $B$ to $A$, the length of the trajectory of the moving point $P$ is
3. $\sqrt{2} \pi$.

As shown in Figure 8, let the intersection point of the semicircle $\odot O$ with the $y$-axis be $M$, and connect $M P$, $M B$, and $M C$. Then
$$
\begin{array}{l} 
\angle B C M \\
= \angle B C A+ \\
\angle A C M \\
= \angle B C A+ \\
\angle A B M \\
= 90^{\circ}+45^{\circ}=135^{\circ} .
\end{array}
$$

Also, $\angle B C P=90^{\circ}$, so
$$
\angle P C M=180^{\circ}-\angle B C M-\angle B C P=135^{\circ} \text {. }
$$

Thus, $\angle B C M=\angle P C M$.
Since $|C B|=|C P|$, we have
$\triangle B C M \cong \triangle P C M$.
Therefore, $|M P|=|M B|=\sqrt{2}$.
Hence, the locus of the moving point $P$ is a semicircle with $M$ as the center and $\sqrt{2}$ as the radius, and its length is $\sqrt{2} \pi$.