3. Rectangle $A B C D$ is a billiard table, with the length-to-width ratio $A B: B C=2: 1$. A ball is shot from the midpoint $P$ of $A B$ at an acute angle $\alpha$ to $A B$ towards side $B C$, and it sequentially bounces off sides $B C$, $C D$, $D A$, and $A B$ before hitting a ball $M$ at the center of the table. Then $\tan \alpha=(\quad)$.
(A) $\frac{3}{4}$
(B) $\frac{5}{8}$
(C) $\frac{7}{12}$
(D) $\frac{9}{16}$
3. B.

As shown in Figure 2, construct the reflection of point $M$ about $AB$ to get point $M_{4}$, the reflection of $M_{4}$ about $AD$ to get point $M_{3}$, the reflection of $M_{3}$ about $CD$ to get point $M_{2}$, and the reflection of $M_{2}$ about $BC$ to get point $M_{1}$.

Let $P M_{1} \cap BC=P_{1}, P_{1} M_{2} \cap CD=P_{2}$, $P_{2} M_{3} \cap DA=P_{3}, P_{3} M_{4} \cap AB=P_{4}$.

According to the principle of reflection, the path $P \rightarrow P_{1} \rightarrow P_{2} \rightarrow P_{3} \rightarrow P_{4} \rightarrow M$ is the trajectory of ball $P$ after it bounces off the edges $BC, CD, DA, AB$ in sequence and hits ball $M$.
Given $BC=a, AB=2a$. By the properties of similar triangles, we have
$$
\begin{array}{l}
\tan \alpha=\frac{BP_{1}}{PB}=\frac{CP_{1}}{CP_{2}}=\frac{DP_{3}}{DP_{2}}=\frac{AP_{3}}{AP_{4}}=\frac{MP}{PP_{4}} \\
=\frac{\left(BP_{1}+CP_{1}\right)+\left(DP_{3}+AP_{3}\right)+MP}{BP+\left(CP_{2}+DP_{2}\right)+\left(AP_{4}+PP_{4}\right)} \\
=\frac{a+a+\frac{a}{2}}{a+2a+a}=\frac{5}{8} .
\end{array}
$$