= Solution
Split the four-<bit> unsigned inputs into two-<bit> halves: $X=4X_H+X_L$ and $Y=4Y_H+Y_L$. Use four copies of the two-<bit> <binary multiplier> to compute $P_{LL}=X_LY_L$, $P_{HL}=X_HY_L$, $P_{LH}=X_LY_H$, and $P_{HH}=X_HY_H$. Then
$$
\boxed{XY=P_{LL}+4P_{HL}+4P_{LH}+16P_{HH}.}
$$
The extreme terms occupy disjoint output positions: $P_{LL}$ uses positions zero to three and $16P_{HH}$ uses positions four to seven. Form $W$ by concatenating their four-<bit> outputs, with $P_{HH}$ in the upper half and $P_{LL}$ in the lower half. This needs wiring alone, so $W=16P_{HH}+P_{LL}$ without an <adder>.
Use the first eight-<bit> <adder> to form $R=P_{HL}+P_{LH}$ after zero-extension, and the second to form $Z=W+(R\ll2)$. Thus \b[four two-<bit> <binary multipliers> and two eight-<bit> <adders> suffice]. The fixed <left bit shift> is also wiring alone.
The middle sum can reach $9+9=18$, so preserve all five of its potentially nonzero <bits>; retaining only four would be wrong. Its shifted value is at most 72. The final result is at most $15\times15=225$, so both additions fit in eight <bits>. Concatenating the disjoint extreme terms is what saves an addition.
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