In this question, conjugation of p-orbitals in the system is misrepresented. It should be noted that, a π-system to be an aromatic or anti-aromatic, there should be a closed loop(s) of p-orbitals (conjugated), which is planner. What makes it an aromatic or anti-aromatic whether it follows Hückel's
rule of (4n+2) π-electrons or not.
Aromatic (most stable): If a planer, closed loop(s) of p-orbital system has (4n+2) π-electrons (in the closed system), it is aromatic. Example:, benzene (a closed p-orbital system with six π-electrons) and naphthalene (a closed p-orbital system with ten π-electrons).
anti-Aromatic (least stable): If a planer, closed loop(s) of p-orbital system has (4n) π-electrons (in the closed system), it is anti-aromatic. Example:, cyclobutadiene (a closed p-orbital system with four π-electrons) and cyclopentadiene carbocation (a closed p-orbital system with four π-electrons).
non-Aromatic (stability depends on the system): Every other systems are non-aromatic. Examples: 5,6-dimethylidenecyclohexa-1,3-diene and heptafulvene (In both cases, each p-orbital system is not closed).
See following diagram: