If X and Y are the inputs, the given circuit works as :

Assume positive-logic levels: LOW = $$0\text{ V}$$, HIGH = $$+5\text{ V}$$. In the given network the anodes of two identical silicon diodes are tied to the input lines $$X$$ and $$Y$$, while their cathodes meet at node $$O$$. Node $$O$$ is pulled up to $$+5\text{ V}$$ through a resistor $$R_L$$ (the load resistor) and is the output.

Step 1 - Input combination $$X = 0$$, $$Y = 0$$
Both anodes are at $$0\text{ V}$$. Each diode is forward-biased (anode $$0\text{ V}$$, cathode initially $$5\text{ V}$$). They conduct and clamp node $$O$$ close to $$0.7\text{ V}$$ (the diode drop). The output is interpreted as LOW.

Step 2 - Input combination $$X = 0$$, $$Y = 1$$
The diode connected to $$X$$ is forward-biased (anode $$0\text{ V}$$), so it again pulls node $$O$$ to about $$0.7\text{ V}$$. The other diode is reverse-biased, but the conducting one is enough to keep the output LOW.

Step 3 - Input combination $$X = 1$$, $$Y = 0$$
This is symmetrical to Step 2. The diode linked with $$Y$$ conducts, forcing the output LOW (≈$$0.7\text{ V}$$).

Step 4 - Input combination $$X = 1$$, $$Y = 1$$
Now both diode anodes and cathodes sit at $$+5\text{ V}$$, so each diode is reverse-biased. No current flows through the diodes. The pull-up resistor $$R_L$$ therefore lifts node $$O$$ to $$+5\text{ V}$$, producing a HIGH output.

Step 5 - Truth table

$$\begin{array}{ccc|c} X & Y & & O \\ \hline 0 & 0 & & 0 \\ 0 & 1 & & 0 \\ 1 & 0 & & 0 \\ 1 & 1 & & 1 \\ \end{array}$$

This truth table is identical to that of a two-input AND gate.

Hence the circuit functions as an AND gate.

Option B which is: AND gate