Two \(4\) bits binary numbers, \(A = 1101\) and \(B = 1010\) are given in the inputs of a logic circuit shown in figure below. The output \((Y)\) will be:

From the given logic circuit, the input \(A\) passes through a NOT gate, giving \(\overline{A}\).

This output \(\overline{A}\) and the input \(B\) are then fed into a NAND gate.

The Boolean expression for the final output \(Y\) is:
\(Y = \overline{\overline{A} \cdot B}\)

Given the \(4\)-bit binary inputs:
\(A = 1101\)
\(B = 1010\)

First, find the bitwise NOT of \(A\):
\(\overline{A} = \overline{1101} = 0010\)

Next, find the bitwise AND of \(\overline{A}\) and \(B\):
\(\overline{A} \cdot B = 0010 \cdot 1010 = 0010\)

Finally, apply the NOT operation for the NAND gate:
\(Y = \overline{0010} = 1101\)

Alternatively, using De Morgan's laws:
\(Y = \overline{\overline{A} \cdot B} = A + \overline{B}\)
\(Y = 1101 + \overline{1010} = 1101 + 0101 = 1101\)

Therefore, the output \(Y\) is \(1101\).

Answer: \(Y = 1101\)