a) Suppose the particle is at the point (x, y). Then
x = a cos(wt)
y = sin(wt)
The acceleration is s'' where s is the displacement vector and each derivative is taken with respect to time. The displacement is a vector from the origin to the point in question, so in terms of the unit vectors i and j:
s = x*i + y*j = a cos(wt)*i + sin(wt)*j
Then taking the derivatives:
s' = -aw sin(wt)*i + w cos(wt)*j
s'' = -aw^2 cos(wt)*i - w^2 sin(wt)*j
Let's play with this a bit and factor a -w^2:
s'' = -w^2 (a cos(wt) + sin(wt)) = -w^2*s
So s'' is a vector that is w^2 longer than s and, most importantly, points in exactly in the opposite direction to s. Since s points from the origin to the point, s'' points from the point to the origin.
b) The magnitude of the acceleration is |s''|, so
|s''| = w^4 * |s|
Thus the two are proportional.
Note carefully that the centripetal acceleration does NOT point toward the origin in general. The centripetal acceleration is the acceleration component in a direction perpendicular to the velocity.