Angular momentum of a single particle moving with constant speed along circular path

From the figure,

An object 's angular momentum is given by 

$\overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{r}}$

where momentum $\overrightarrow{\mathrm{p}}=\mathrm{m}\overrightarrow{\mathrm{v}}$

The velocity of the particle will be always tangential to the circular path and the radius vector will be radial. 

the angular momentum which is a cross product of these two will be perpendicular to the plane of the path. Also, the magnitude does not change as long as the velocity remains constant. 

Hence the correct answer is remains same in magnitude and direction.