K4 and Q3 are graphs with the following structures. Which one of the following statements is TRUE in relation to these graphs?

A graph is planar if it can be drawn in a plane without any of its edges crossing.

1. Graph $K_4$ (complete graph on 4 vertices): It has $V = 4$ vertices and $E = 6$ edges. $K_4$ can be drawn in the plane without edge crossings by placing one vertex inside the triangle formed by the other three and connecting it to all three vertices. Hence, $K_4$ is planar.

2. Graph $Q_3$ (3-dimensional hypercube / 3-cube graph): It has $V = 8$ vertices and $E = 12$ edges. It is isomorphic to the 3D cube graph, which can be drawn as a planar graph (the standard Schlegel diagram of a cube: an inner square inside an outer square with corresponding vertices connected). Since it contains neither $K_5$ nor $K_{3,3}$ as a minor/subdivision, by Kuratowski's theorem, $Q_3$ is planar.

Therefore, Both $K_4$ and $Q_3$ are planar, making (B) the correct choice.