The ionization energy of an element is the energy required to remove an electron from a gaseous atom or ion. The higher the ionization energy, the more difficult it is to remove an electron. Ionization energy tends to increase across a period on the periodic table and decreases down a group. This pattern is due primarily to the increasing nuclear charge and the decreasing radius of atoms across a period, which results in a stronger attraction between the nucleus and the outermost electron.

Let's analyze the given options based on their electron configurations and how they fit into the periodic table:

• Option A: [Ne] 3s²3p¹ - This configuration corresponds to aluminum (Al), which is in the 3rd group and 3rd period of the periodic table.
• Option B: [Ne] 3s²3p³ - This configuration corresponds to phosphorus (P), which is in the 5th group and 3rd period.
• Option C: [Ne] 3s²3p² - This configuration corresponds to silicon (Si), which is in the 4th group and 3rd period.
• Option D: [Ne] 3d¹⁰4s²4p³ - This configuration corresponds to arsenic (As), which is in the 5th group but in the 4th period.

Comparing the elements within the same period (options A, B, and C), phosphorus (Option B) has electrons filled up to the higher p orbital (3p³) which leads to a higher effective nuclear charge exerted on the valence electrons compared to aluminum (3p¹) and silicon (3p²). This increased effective nuclear charge results in higher electron-nucleus attraction, thus higher ionization energy. On the other hand, arsenic (Option D), though having an increased nuclear charge, is located in a higher period where increased shielding effect and a larger atomic radius might reduce the ionization energy compared to elements in the 3rd period.

Therefore, amongst the given elements, phosphorus (Option B: [Ne] 3s²3p³) has the highest ionization energy because it has a higher effective nuclear charge compared to the others in the same period and less shielding effect than arsenic which is in the 4th period.