Ans: A
Note that the slope of the straight line $2x+(a+3)y-5=0$ is $\dfrac{-2}{a+3}$.
Note also that the slope of the straight line $ax-4y+1=0$ is $\dfrac{a}{4}$.
Since the two straight lines are perpendicular to each other, we have
$\begin{array}{rcl}
\dfrac{-2}{a+3} \times \dfrac{a}{4} & = & -1 \\
-2a & = & -4(a+3) \\
-2a & = & -4a-12\\
2a & = & -12 \\
a & = & -6
\end{array}$
