The correct answer is $\boxed{\text{B) 9°}}$.
The maximum range of a projectile is achieved when the angle of projection is $\theta = \arctan \left(\frac{v_y}{v_x}\right)$, where $v_y$ is the vertical component of the initial velocity and $v_x$ is the horizontal component of the initial velocity. In this case, $v_y = 60 \sin 9^\circ = 5.196 m/s$ and $v_x = 60 \cos 9^\circ = 58.778 m/s$, so the angle of projection is $\theta = \arctan \left(\frac{5.196}{58.778}\right) = 8.99^\circ$. However, the tunnel is only 554 cm high, so the particle will hit the ceiling if the angle of projection is greater than $\arcsin \left(\frac{554}{60}\right) = 10.3^\circ$. Therefore, the maximum angle of projection for which the particle will not hit the ceiling is $\boxed{9^\circ}$.
The other options are incorrect because they are not the maximum angle of projection for which the particle will not hit the ceiling.