Take the molecular (internuclear) axis along $z$.

(A) ${\sigma }_{2p_z}$ formed by end-on overlap of two 2p${}_z$ lobes has two nodal planes containing the axis (the $xz$- and $yz$-planes are not nodal here; rather the angular nodes from each 2p orbital contribute two planar nodes perpendicular to the orbital axes that the lobes lie along). Each constituent 2p${}_z$ has one nodal plane perpendicular to its lobe; in the bonding combination two nodal surfaces persist. Correct.

(B) ${\sigma }_{2p_z}^{\ast }$ has an additional node between the two nuclei (perpendicular to the bond axis), giving three nodal planes total. The single antibonding node is the $xy$-plane between the nuclei, not the $xz$-plane along the axis. Wrong.

(C) ${\pi }_{2p}$ orbitals (formed from $p_x$ or $p_y$) have one nodal plane containing the molecular axis (the plane perpendicular to the lobes). They do not have a node perpendicular to the axis through the centre. Wrong.

(D) ${\pi }_{2p}^{\ast }$ has two nodal planes: the one containing the molecular axis (inherited from $\pi$) and an additional perpendicular one between the nuclei. One of these planes contains the molecular axis. If the $\pi$ orbitals are formed from $p_x$, the nodal plane containing the axis is the $yz$-plane; for $p_y$ overlap it is the $xz$-plane. The statement that ${\pi }^{\ast }$ has a nodal plane containing the molecular axis is correct. Correct.