what is difference between twist and classcial velocity

In ch 8.2.1 Classical Formulation in the book Modern Robotics by Frank C. Park I know some remarks about this formula.The origin of coordinate system is coincidence with the mass center of the rigid body. $$w_b$$ is the angular velocity of the body. It is noticeable that $$v_b$$ is not the velocity of one particular fixed point in the body, but is the velocity of one selected point in the body and the point is in coincidence with the origin of the system instantaneously. Based on these, we can use the formula compute arbitray point's velocity and acceleration. The point's coordiante in the system is $$p_i$$. The $$p_i$$ rightly reprensent one concrete point in the body.

Now question is that , if I take $$v_b$$ as the fixed point(the mass center) in the body , how can I get velocity and acceleration of arbitray point in the body? What is the formula ? By this I think I can understand the different between twist and the classical velocity

In these equations from Modern Robotics (by Park and Lynch), the fixed inertial frame $${b}$$ is both the reference frame used to define all of the coordinate vectors and has its origin located at the centre of mass (COM) of the body.
The body's angular velocity relative to $$\{b\}$$ is $$\omega_b$$, as you say, and is a property of the body which is dependent only on the orientation of the frame and not on the location of its origin. The variable $$v_b$$ is the linear velocity of the centre of mass of the body as it moves past the (instantaneously fixed, inertial) frame $$\{b\}$$, and is dependent on both the origin and orientation of the frame.
You have also already answered your own question, as the way to find out the linear velocity and acceleration of any single point $$p_i$$ (which is rigidly attached to the body) relative to the frame $$\{b\}$$ is to compute the equations you have shared above:
• The linear velocity of the point (i.e., $$\dot p_i$$) is simply $$\dot p_i = v_b + \omega_b \times p_i$$, which combines the linear velocity of the COM ($$v_b$$) and the linear velocity of the point caused by the rotational velocity of the body ($$\omega_b$$) acting on the perpendicular distances the point is away from the COM in each direction ($$\times p_i$$).
• The linear acceleration of the point (i.e., $$\ddot p_i$$) is the derivative of the linear velocity, which for brevity's sake I won't retype but will refer you to the image in your initial question. This equation combines the linear acceleration of the COM ($$\dot v_b$$), the linear acceleration of the point caused by the rotational acceleration of the body ($$\dot \omega_b$$) acting on the perpendicular distances the point is away from the COM in each direction ($$\times p_i$$), and the velocity product $$\omega_b \times \dot p_i = \omega_b \times (v_b + \omega_b \times p_i)$$ which includes both the centripetal and Coriolis accelerations of the body.