Let $D$ be the distance $AB$ and let the speeds of $P, Q$ be $v_P, v_Q$.

At the first meeting, $P$ has covered $7.5$ and $Q$ has covered $D-7.5$, both in the same time. So

$

\frac{v_P}{v_Q}=\frac{7.5}{D-7.5}. \tag{1}

$

By the time of the second meeting, both cyclists have reached the far town and turned back (this must be checked at the end). Between start and second meeting, the two together cover $3D$ (the first meeting accounts for a combined $D$; after that, until they meet again, they together cover an additional $2D$).

In total, $P$ travels $D$ to reach $B$, then turns and is at $4$ km from $B$, so $P$'s total distance is $D + 4$. Hence $Q$'s total distance is $3D - (D+4) = 2D - 4$.

$

\frac{v_P}{v_Q}=\frac{D+4}{2D-4}. \tag{2}

$

Equating (1) and (2):

$

\frac{7.5}{D-7.5}=\frac{D+4}{2D-4} ;\Longrightarrow; 7.5(2D-4)=(D+4)(D-7.5).

$

$15D-30 = D^2 - 3.5D - 30 \;\Longrightarrow\; D^2 - 18.5D = 0 \;\Longrightarrow\; D = 18.5.$

Verification of the ``both have turned'' assumption: With $D=18.5$, speed ratio $v_P:v_Q = 7.5:11 = 15:22$. Time for $P$ to reach $B$ is $18.5/15 \approx 1.233$; time for $Q$ to reach $A$ is $18.5/22 \approx 0.841$. Time of second meeting solves $37 - 15t = 22t - 18.5$, giving $t=1.5$, which exceeds both turnaround times. Position $= 14.5$, i.e., $4$ from $B$. \checkmark

Why this is CAT-level: The natural first approach (``ratio of distances at first meeting and total covered $=3D$'') is correct only if both cyclists have turned by the second meeting -- a hidden case-check that distinguishes a percentile-pushing solver.

Answer: (B) $18.5$.