= Solution
In the original rest frame each leg takes $d/c$, so the round trip takes $2d/c$. For the moving frame, use the coordinate differences between the same three events. The <Lorentz transformation> gives
$$
\boxed{\Delta t'_{AB}=\frac{\gamma d}{c}(1-\beta),\quad\Delta t'_{BC}=\frac{\gamma d}{c}(1+\beta),\quad\Delta t'_{AC}=\frac{2\gamma d}{c}.}
$$
The outbound leg is shorter because the mirror approaches the outgoing ray; the inbound leg is longer because the source retreats from the returning ray. Their sum is larger than the original round-trip <proper time> by the <time dilation> factor $\gamma$.
These are coordinate times for return to the original source, not a claim that the pulse returns to the second observer's body at event $C$. If that observer starts at $A$, it intercepts the returning light at a different event, earlier than $C$, at its own time $2\gamma d(1-\beta)/c$. Distinguishing those events is necessary when interpreting “seen” or “return”.
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