K-map ΠM(0,2,4,6,7,8,10,13,15) and 8:1 MUX

$Y(A,B,C,D) = \Pi M(0,2,4,6,7,8,10,13,15)$ means Y = 0 at these cells. So Y = 1 at the remaining cells:

$$Y = \Sigma m(1,3,5,9,11,12,14)$$

K-map simplification

        CD
AB      00  01  11  10
      +---+---+---+---+
 00   | 0 | 1 | 1 | 0 |
      +---+---+---+---+
 01   | 0 | 1 | 0 | 0 |
      +---+---+---+---+
 11   | 1 | 0 | 0 | 1 |
      +---+---+---+---+
 10   | 0 | 1 | 1 | 0 |
      +---+---+---+---+

Grouping the 1s:

• Quad m(1,3,9,11) → $\overline{B}D$
• Pair m(1,5) → $\overline{A}\,\overline{C}D$
• Pair m(12,14) → $AB\overline{D}$

$$Y = B'D + A'C'D + ABD'$$

(Grouping the 0s gives the minimal POS $Y = (A+D)(B+D)(\overline{A}+\overline{B}+\overline{D})(\overline{B}+\overline{C}+\overline{D})$, which has more terms, so the SOP form is used.)

Realization with an 8:1 multiplexer

Use A, B, C as the select lines $S_2 S_1 S_0$ and connect each data input $I_0$–$I_7$ to 0, 1, D or $\overline{D}$. For each value of ABC, compare Y for D = 0 and D = 1:

          +-----------+
 D  ───── | I0        |
 D  ───── | I1        |
 D  ───── | I2        |
 0  ───── | I3  8:1   |── Y
 D  ───── | I4  MUX   |
 D  ───── | I5        |
 D' ───── | I6        |
 D' ───── | I7        |
          +-----------+
           S2  S1  S0
           A   B   C

 D ──[NOT]── D'

Only one NOT gate is needed besides the multiplexer. Check: for $ABC = 110$, Y = $\overline{D}$, so minterm 12 (D=0) gives 1 and minterm 13 gives 0, as required.