Given equation of SHM

$x=\sin\left[\pi\left(t+\frac13\right)\right] \text{ m}$

This is of the form: $x=A\sin(\omega t+\phi)$

So, $A=1\text{ m}, \qquad \omega=\pi\text{ rad s}^{-1}$

Find velocity equation

Velocity is: $v=\frac{dx}{dt}$

Differentiate: $v=\pi\cos\left[\pi\left(t+\frac13\right)\right] \text{ m s}^{-1}$

Evaluate at $t=1\,$s

$v=\pi\cos\left[\pi\left(1+\frac13\right)\right]$ $=\pi\cos\left(\frac{4\pi}{3}\right)$

Now, $\cos\left(\frac{4\pi}{3}\right)= -\frac12$

Hence, $v=\pi\left(-\frac12\right)=-\frac{\pi}{2}\text{ m s}^{-1}$

Find speed

Speed is magnitude of velocity: $|v|=\frac{\pi}{2}\text{ m s}^{-1}$

Using $\pi=3.14$, $|v|=\frac{3.14}{2}=1.57\text{ m s}^{-1}$

Convert to cm/s

$1.57\text{ m s}^{-1}=157\text{ cm s}^{-1}$

Match with options

$\boxed{157\text{ cm s}^{-1}}$

So the correct option is B.