For the given circuit (shown in part (A)) the time dependent input...

Given: Input is an alternating voltage with peak values $+20$ V and $-20$ V. Output is clipped at $+5$ V and $-5$ V.

The existing branch (given in circuit) contains an ideal diode (anode down) in series with a Zener diode ($V_z = 5$ V).

During negative half cycle: The ideal diode is forward-biased, and the Zener diode is reverse-biased. When $|v_{in}| > 5$ V, the Zener breaks down, clipping the output at $-5$ V. So the existing branch handles the negative half cycle.

During positive half cycle: The ideal diode in the existing branch is reverse-biased, so that branch is open. The branch between X and Y must clip the output at $+5$ V.

For clipping at $+5$ V during the positive half cycle, the XY branch must have an ideal diode (anode towards X) in series with a Zener diode (cathode towards X, $V_z = 5$ V).

When $v_{in} > +5$ V: ideal diode is forward-biased, Zener is reverse-biased and clamps at $5$ V $\Rightarrow$ output clipped at $+5$ V.
When $v_{in}$ is negative: ideal diode is reverse-biased $\Rightarrow$ branch is open, leaving the existing branch to handle clipping.

This matches Option B.