Let us suppose the initial output of all the JK flip flops is $1$
So we can draw the below table to get the output $Q_3$
$\small \begin{array} {c|c|c|c|c|c|c|c|c|c|c|c|c|c } \hline \text{Index} & Q_3 & Q_2 & Q_1 &Q_0 & J_3 & K_3 & J_2 & K_2 & J_1 & K_1 & J_0 & K_0 \\\hline & & & & & Q_2 & Q_2’ & Q_1 & Q_1’ & Q_0 & Q_0’ & Q_3 \oplus Q_2 & J_0’ \\\hline 0 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\\hline 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 1 \\\hline 2 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1 \\\hline 3 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 \\\hline 4 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 \\\hline 5 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 \\\hline 6 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \\\hline 7 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 \\\hline 8 & 0 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\\hline 9 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 \\\hline 10 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \\\hline 11 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 \\\hline 12 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\\hline 13 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 \\\hline 14 & 0 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\\hline & 1 & 1 & 1 & 1 \\\hline \end{array}$
From the above table $Q_3$ that is output is $1111\; 0001\; 0011\; 0101$
So the answer should be C.