Number of ways to choose path from $A$ to $B$ = 4C1
Number of ways to choose path from $B$ to $C$= 3C1
Now during roundtrip
Number of ways to choose path from $C$ to $B$ = 3C1
Number of ways to choose path from $B$ to $A$= 4C1
Total possibe combinations= $4C1 * 3C1 * 3C1 * 4C1 $= $4*3*3*4= 144$
Hence Option C) is correct answer