5. Given positive numbers $a, b, c, d$ satisfy $a+2 b=1, c+2 d=1$, then the minimum value of $\frac{1}{a}+\frac{1}{b c d}$ is
A. $\frac{1}{16}$
B. 8
C. $\frac{1}{5}$
D. 25
Solve $2 c d \leqslant \frac{1}{4} \Rightarrow \frac{1}{a}+\frac{1}{b c d} \geqslant \frac{1}{a}+\frac{8}{b}=\frac{1}{a}+\frac{16}{2 b} \geqslant \frac{(1+4)^{2}}{a+2 b}=25$, so the answer is $D$.