For the given parallelogram, the diagonals AC and BD intersect at point E. That means;
[tex]\begin{gathered} BE=DE \\ \text{Also,} \\ AE=CE \end{gathered}[/tex]Substitute for the given values and we'll have;
[tex]\begin{gathered} BE=DE \\ y+10=4y-8 \\ \text{Collect all like terms and you'll have;} \\ y-4y=-8-10 \\ -3y=-18 \end{gathered}[/tex]Divide both sides by -3 and you'll have;
[tex]\begin{gathered} \frac{-3y}{-3}=\frac{-18}{-3} \\ y=6 \end{gathered}[/tex]Note that line segment BD is made up of;
[tex]\begin{gathered} BD=BE+DE \\ BD=(y+10)+(4y-8)_{} \\ BD=y+10+4y-8 \\ BD=4y+y+10-8 \\ BD=5y+2 \\ \text{When y=6, we now have} \\ BD=5(6)+2 \\ BD=30+2 \\ \text{BD}=32 \end{gathered}[/tex]Therefore, the answer is option B
