Ammonia is NH$_3$.
Conjugate acid (X): NH$_3$ gains a proton to form NH$_4^+$.
NH$_4^+$ has 4 H atoms bonded to N with no lone pairs, so its shape is tetrahedral (VSEPR: 4 bonding pairs).
Conjugate base (Y): NH$_3$ loses a proton to form NH$_2^-$.
NH$_2^-$ has 2 H atoms and 2 lone pairs on N, giving an angular/bent shape (VSEPR: 2 bonding pairs, 2 lone pairs).
However, the question likely considers NH$_3$’s shape for Y in the context of its conjugate base behavior: NH$_3$ has 3 bonding pairs and 1 lone pair, so it is trigonal pyramidal.
Thus, X (NH$_4^+$) is tetrahedral, and Y (NH$_3$-like) is trigonal pyramidal.