A graph is planar if it can be redrawn in the plane with no edges crossing. By Kuratowski's theorem a graph is non-planar exactly when it contains a subdivision of K₅ or K_3,3; for a connected planar simple graph, e ≤ 3v − 6.

Step 1 — Graph A has 6 vertices and 8 edges: the cycle a–b–d–e–f–c–a plus the chords b–f and c–d. The two chords cross only because both are drawn inside the cycle; drawing c–d outside the cycle removes the crossing, so A is planar.

Step 2 — Graph B has 6 vertices and 10 edges: a–c, a–d, a–e, a–f, b–c, b–d, b–f, c–e, d–f, e–f. Without a–f it is the triangular prism with triangles a–c–e and b–d–f joined by a–d, c–b, e–f, which is planar; a–f is a diagonal of the quadrilateral face a–e–f–d, so it can be added without a crossing. B is planar.

Step 3 — Graph C has 9 vertices and 15 edges. Take the parts {a, e, f} and {b, d, g}: a–b, a–d, a–g, e–b, e–d, f–b, f–g are edges, e reaches g through the path e–h–g and f reaches d through f–c–d. This is a subdivision of K_3,3, so by Kuratowski's theorem C is not planar.

So the planar graphs are A and B. Any choice that includes C is wrong, and the choices naming only A or only B each leave out one planar graph.

A and B only option is correct.