To find the displacement, we first integrate the acceleration to find the velocity. Given:

\(a(t) = 0.075t^2 - 1.5t + 5\)

Integrate \(a(t)\) to find \(v(t)\):

\(v(t) = \int (0.075t^2 - 1.5t + 5) \, dt = 0.025t^3 - 0.75t^2 + 5t + C\)

Since the particle starts from rest, \(v(0) = 0\), so \(C = 0\).

Thus, \(v(t) = 0.025t^3 - 0.75t^2 + 5t\).

Integrate \(v(t)\) to find \(s(t)\):

\(s(t) = \int (0.025t^3 - 0.75t^2 + 5t) \, dt = 0.00625t^4 - 0.25t^3 + 2.5t^2 + C\)

Since the displacement is zero at \(t = 0\), \(C = 0\).

Thus, \(s(t) = 0.00625t^4 - 0.25t^3 + 2.5t^2\).

To find when \(s(t) = 0\) again, solve:

\(0.00625t^4 - 0.25t^3 + 2.5t^2 = 0\)

Factor out \(t^2\):

\(t^2(0.00625t^2 - 0.25t + 2.5) = 0\)

\(t^2 = 0\) gives \(t = 0\), and solving \(0.00625t^2 - 0.25t + 2.5 = 0\) gives \(t = 20\).

Therefore, the time taken for P to return to the point O is 20 seconds.