Sound waves of frequency 660 Hz fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particles have maximum amplitude of vibration is  $\frac{P}{16}$  meters. Then P will be (Speed of sound 330 m/s)

Here, sound waves fall normally on a perfectly reflecting wall, hence the incident and reflected wave will form the stationary wave=form, since there is no displacement of particles at wall hence this point is node.
As the separation between two nodes is  $\frac{\lambda }{2}m$  the antinode i.e the point of displacement of particles is at a distance.
$\frac{\frac{\lambda }{2}}{2}=\frac{\lambda }{4}m$  from wall.
Now, we know
$\lambda =\frac{v}{m}$ 
 $\lambda =\frac{330}{660}\mathrm{............}\left(given\right)$
  $\lambda =\frac{1}{2}m$
Hence, the distance of anti-node from wall is  $\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}m$
Therefore the shortest distance from the wall at which the air particle has maximum amplitude of vibration (ant0-node)is $\frac{1}{8}m$  i.e 0.125m