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Correct answer: The hybridisation of M is sp3
Pyramidal MX3 means M has 3 bonding pairs and 1 lone pair => sp3 hybridised (tetrahedral electron geometry). In p(pi)-d(pi) bonding: X contributes its filled p orbital and M accepts using its empty d orbital (back bonding). So M uses its d-orbitals, not X. This rules out option C. No hypervalency (not more than 4 electron pairs in total) — option A (hypovalent, meaning fewer bonds) is wrong. Two lone pairs on M would give a different geometry (T-shaped for AX3E2) — option B is wrong. Option D (sp3 for M) is correct.