Example 6 As shown in the figure, two rivers converge at point $O$, forming a $75^{\circ}$ angle. The traveler lives at point $P$, 200 meters from point $O$, and 100 meters from the riverbank $AO$. He wishes to go to any point $C$ on the riverbank $AO$ to enjoy the scenery, then turn to any point $D$ on the riverbank $BO$ to admire the view, and finally return to his residence. What is the minimum distance the traveler has to walk? $\qquad$ meters.
(Fourth Wu Yang Cup Junior High School Mathematics Competition)
Solution: Make the symmetric points of point $P$ about $AO$ and $BO$ as $P'$ and $P''$, respectively. Then, $PC + CD + DP = P'C + CD + DP''$ is the length of the broken line between $P'$ and $P''$. To minimize this, $P'P''$ must be a straight line segment.
From the given, we have $\angle AOP = 30^\circ$.

Also, $\angle AOB = 75^\circ$, so $\angle POB = 45^\circ$.
Thus, $\angle P'OP'' = 150^\circ$.
$$
\begin{aligned}
\text{Also, } OP' = OP'' = & OP = 200 \text{ meters,} \\
\text{Therefore, } (P'P'')^2 = & (OP'')^2 + (OP')^2 \\
& - 2 \cdot OP' \cdot OP'' \cdot \cos 150^\circ \\
= & 10000(8 + 4\sqrt{3}) \\
= & 10000(\sqrt{6} + \sqrt{2})^2.
\end{aligned}
$$

Therefore, $P'P'' = 100(\sqrt{6} + \sqrt{2})$.
Hence, the traveler needs to walk at least $100(\sqrt{6} + \sqrt{2})$ meters.