1. Identify the elements from the clues

   We are given a molecule of type $AX_4Y$, and all elements belong to the p-block.

   • Element $X$ has the highest electronegativity among all known elements.

     Therefore, $X = \mathrm{F}$.

   • Element $Y$ has the second highest electronegativity among all known elements.

     Therefore, $Y = \mathrm{O}$.

   • Element $A$ is:

     • rarest,
     • monoatomic,
     • non-radioactive from its group,
     • and has the lowest ionization enthalpy among $A, X, Y$.

     The p-block element fitting “rarest, monoatomic, non-radioactive from its group” is xenon, $\mathrm{Xe}$, from group 18.

   Hence the molecule is: $\mathrm{XeF_4O}$ which is commonly written as $\mathrm{XeOF_4}$.

2. Determine the central atom electron count

   Central atom: $\mathrm{Xe}$

   Valence electrons on Xe = $8$.

   In $\mathrm{XeOF_4}$:

   • Xe forms 4 bonds with F
   • Xe forms 1 bond with O (effectively a multiple bond in resonance description, but for VSEPR it counts as one electron domain)

   Thus total bonded atoms around Xe = $5$.

3. Count electron domains using VSEPR

   For $\mathrm{XeOF_4}$, xenon has:

   • $5$ bond pairs (4 Xe–F and 1 Xe–O domain)
   • $1$ lone pair

   So steric number = $5 + 1 = 6$

   Electron-pair geometry is therefore octahedral.

4. Determine the molecular shape

   With steric number $6$ and one lone pair, the molecular shape is obtained from octahedral geometry by removing one position for the lone pair.

   This gives the shape: $\boxed{\text{square pyramidal}}$

   In $\mathrm{XeOF_4}$, the lone pair and the $\mathrm{Xe=O}$ domain prefer opposite positions, leaving four F atoms in one square plane and O at the apex.

5. Check options

   • A: Pentagonal planar $\to$ incorrect
   • B: Square pyramidal $\to$ correct
   • C: Trigonal bipyramidal $\to$ incorrect
   • D: Octahedral $\to$ electron-pair geometry, not molecular shape

Therefore, the correct option is: $\boxed{\text{B: Square pyramidal}}$