3. As shown in Figure 2, given a regular tetrahedron $A B C D$ with edge length 1, $M$ is the midpoint of $A C$, and $P$ lies on the line segment $D M$. Then the minimum value of $A P+B P$ is
3. $\sqrt{1+\frac{\sqrt{6}}{3}}$.

Let $\angle B D M=\theta$. In $\triangle B D M$,
$$
B D=1, B M=M D=\frac{\sqrt{3}}{2} \text {. }
$$

Thus, $\cos \theta=\frac{\sqrt{3}}{3}, \sin \theta=\frac{\sqrt{6}}{3}$.
As shown in Figure 5, rotate $\triangle B D M$ around $D M$ so that $\triangle B D M$ lies in the plane $A C D$, at this point $B$ is at $B^{\prime}$. Connect $A B^{\prime}$ and $B^{\prime} P$. The minimum value sought is the length of $A B^{\prime}$. It is easy to see that
$$
\angle A D B^{\prime}=\theta+30^{\circ} \text {. }
$$

Therefore, $A B^{\prime 2}=A D^{2}+D B^{\prime 2}-2 A D \cdot D B^{\prime} \cos \angle A D B^{\prime}$
$$
\begin{array}{l}
=1^{2}+1^{2}-2 \cos \left(\theta+30^{\circ}\right) \\
=2-2\left(\cos \theta \cdot \cos 30^{\circ}-\sin \theta \cdot \sin 30^{\circ}\right) \\
=1+\frac{\sqrt{6}}{3} .
\end{array}
$$

Thus, $A B^{\prime}=\sqrt{1+\frac{\sqrt{6}}{3}}$.