The logic operations performed by the given digital circuit is equivalent to:

Let the two external inputs be labelled as $$A$$ and $$B$$.

Step 1 - Identify each block in the diagram.
  • Each small triangle with a bubble at its output is an INVERTER (NOT gate).
  • The larger gate whose symbol has a curved input side and a bubble at its output is a NOR gate.

Step 2 - Write the Boolean expression of every stage.
  • Output of the first inverter is $$\overline{A}$$.
  • Output of the second inverter is $$\overline{B}$$.
  • The NOR gate receives $$\overline{A}$$ and $$\overline{B}$$, so its output $$Y$$ is
    $$Y = \overline{\overline{A} + \overline{B}} \quad -(1)$$

Step 3 - Simplify the expression using De Morgan’s theorem.
  De Morgan’s theorem states
  $$\overline{X + Z} = \overline{X} \cdot \overline{Z}$$
  Applying this to $$(1)$$ with $$X = \overline{A}$$ and $$Z = \overline{B}$$ gives
  $$Y = \overline{\overline{A}} \cdot \overline{\overline{B}}$$

Since a double negation cancels, $$\overline{\overline{A}} = A$$ and $$\overline{\overline{B}} = B$$. Hence
  $$Y = A \cdot B$$

Step 4 - Interpret the final Boolean result.
  The output expression $$Y = A \cdot B$$ is the logical AND of the inputs.

Therefore, the overall operation performed by the given circuit is an AND gate.
Option B (AND) is correct.