Understand the potential

The particle moves in a potential of the type $V(x)=\alpha x^4 \quad (\alpha>0)$ near the origin, and for large enough $|x|$, the potential becomes a constant value $V_0$.

So the potential is like a well around $x=0$, and outside some region it flattens to the constant value $V_0$.

Condition for periodic motion

For a particle to execute periodic motion, it must remain bound between two turning points.

That means its total energy $E$ must be such that there exist points where $E=V(x),$ and beyond those points the particle cannot go because kinetic energy would become negative.

Since kinetic energy is $T=E-V(x),$ physically allowed motion requires $E\ge V(x).$

Check different energy ranges

Case 1: $E<0$

Here, near the origin, $V(x)=\alpha x^4\ge 0.$ Also outside, $V(x)=V_0$ which is also above the minimum shown in the figure. Thus $V(x)\ge 0$ everywhere.

If $E<0$, then $E<V(x) \quad \text{for all } x,$ so no motion is possible.

Hence, A is false.

Case 2: $E>0$

This alone is not sufficient for periodic motion. If $E$ becomes larger than or equal to the outer constant level $V_0$, the particle can escape to the flat region and will not remain bound.

So B is not always true.

Case 3: $0<E<V_0$

In this case, near the origin the particle has allowed motion because $V(x)=\alpha x^4$ can be less than $E$.

Turning points are given by $\alpha x^4=E$ $x=\pm \left(\frac{E}{\alpha}\right)^{1/4}.$

Since $E<V_0$, the particle cannot reach the outer flat region where the potential is $V_0$.

Therefore it remains trapped between two turning points and performs periodic motion.

Hence, C is true.

Case 4: $E>V_0$

Then in the outer region where $V(x)=V_0$, $T=E-V_0>0.$ So the particle can move off to infinity with nonzero kinetic energy. The motion is not bound, hence not periodic.

Therefore, D is false.

Final conclusion

The particle performs periodic motion only when $V_0>E>0.$

So the correct option is C.