Step 1: A quicker way to reach the same answer is to first count how many orbitals copper needs. The complex has 4 ligand nitrogen atoms, each donating one lone pair, so copper needs exactly 4 empty hybrid orbitals to accept these 4 lone pairs, ruling out sp3d (5 orbitals) and sp3d2-type options immediately, and narrowing the choice to the 4-orbital hybridizations among the options given, $sp^3$ and $dsp^2$.
Step 2: Now decide between $sp^3$ (tetrahedral) and $dsp^2$ (square planar) using the metal's d-electron count. Copper here is present as $Cu^{2+}$, since the ligand $NH_3$ is neutral and the complex charge (2+) must come entirely from the metal. $Cu^{2+}$ has the configuration $[Ar]3d^9$.
Step 3: A $d^9$ ion has 9 electrons spread over 5 d orbitals, meaning only one d orbital is left available to take part in bonding while the metal still keeps its other 4d orbitals filled with the remaining electrons. This single available d orbital combines with one s and two p orbitals, not three p orbitals, so the hybridization must be $dsp^2$, not $sp^3$. This also matches the well known square planar geometry seen for $Cu^{2+}$ ammine complexes.
\[\boxed{dsp^2 \text{ hybridization, square planar shape}}\]