The force on a charged particle in magnitude field

$F=q\left(v\times B\right)$

The proton is moving along the axis of solenoid. Magnetic field inside the solenoid is parallel to the axis of solenoid.

$\therefore \overrightarrow{v}\parallel \overrightarrow{B}$

$F=qvB\sin \theta$

$\theta =0$

$F=0$

$\therefore$ The proton will continue to move with velocity along the axis.