Consider the two cascaded 2-to-1 multiplexers as shown in the figure. The minimal sum of products form of the output $X$ is

Let us analyze the output of each 2-to-1 multiplexer step by step:

1. First (left) Multiplexer:

   • Inputs: $I_0 = 0$, $I_1 = R$

   • Select line: $S = P$

   • Output:

     $$\text{Out}_1 = \bar{P} \cdot I_0 + P \cdot I_1 = \bar{P} \cdot 0 + P \cdot R = PR$$

2. Second (right) Multiplexer:

   • Inputs: $I_0 = \bar{R}$, $I_1 = \text{Out}_1 = PR$

   • Select line: $S = Q$

   • Output $X$:

     $$X = \bar{Q} \cdot I_0 + Q \cdot I_1 = \bar{Q}\bar{R} + Q(PR) = \bar{Q}\bar{R} + PQR$$

3. Minimality Check:

   • The minterms are $m(0, 1)$ with $Q=0, R=0$ and $m(7)$ with $P=1, Q=1, R=1$.

   • The terms $\bar{Q}\bar{R}$ and $PQR$ cannot be simplified further.

   • Thus, the minimal Sum of Products (SOP) expression is $\bar{Q}\bar{R} + PQR$.

Therefore, the correct option is (D).