We know that the general equation for the output($Z$) of an $8 \times 1 \ Mux$ is
\[ \large Z = \sum_{k=0}^{7} m_kI_k \]
where $m_k$ is the minterm of the $3$ select variables (here $A, B, C$) and $I_k$ is the corresponding data input.
$\begin{align*}
f(A,B,C,D) &= \prod (1,5,12,15) \\
\\ &= \sum(0,2,3,4,6,7,8,9,10,11,13,14) = \sum_{k=0}^{7} m_kI_k \\
\\ &= \sum(\underbrace{\mathbf{000}0}_{\underset{\LARGE{I_0 = \bar D}}{\large{m_0\bar D}}},
\underbrace{\mathbf{001}0, \mathbf{001}1}_{\underset{\LARGE{I_1 = 1}}{\large{m_1(\bar D+D)}}},
\underbrace{\mathbf{010}0}_{\underset{\LARGE{I_2 = \bar D}}{\large{m_2\bar D}}},
\underbrace{\mathbf{011}0, \mathbf{011}1}_{\underset{\LARGE{I_3 = 1}}{\large{m_3(\bar D+D)}}},
\underbrace{\mathbf{100}0, \mathbf{100}1}_{\underset{\LARGE{I_4 = 1}}{\large{m_4(\bar D+D)}}},
\underbrace{\mathbf{101}0, \mathbf{101}1}_{\underset{\LARGE{I_5 = 1}}{\large{m_5(\bar D+D)}}},
\underbrace{\mathbf{110}1}_{\underset{\LARGE{I_6 = D}}{\large{m_6 D}}},
\underbrace{\mathbf{111}0}_{\underset{\LARGE{I_7 = \bar D}}{\large{m_7\bar D}}}) \\
\end{align*}$
$\therefore Ans\ - B.$
Here the shorter way would be to directly find out the four pins for which output of $f$ will depend on $D$ by the $CPOS$ form given, since there are only four inputs for which $f$ is $0$.