For a particle in uniform circular motion the acceleration $$\vec{a}$$ at a point $$P(R, \theta)$$ on the circle of radius R is (here $$\theta$$ is measured from the $$x$$-axis)

The position vector of a particle at point $$P(R,\theta)$$ on a circle of radius $$R$$ is

$$\vec r = R\cos\theta \,\hat i + R\sin\theta \,\hat j$$

In uniform circular motion the angular speed $$\omega$$ is constant, so $$\theta = \omega t$$.

Velocity is the time-derivative of the position vector:

$$\vec v = \frac{d\vec r}{dt} = \frac{d\theta}{dt}\,\frac{d\vec r}{d\theta} = \omega\!\left(-R\sin\theta\,\hat i + R\cos\theta\,\hat j\right) = -R\omega\sin\theta\,\hat i + R\omega\cos\theta\,\hat j$$

Acceleration is the time-derivative of velocity:

$$\vec a = \frac{d\vec v}{dt} = \omega\!\left(-R\omega\cos\theta\,\hat i - R\omega\sin\theta\,\hat j\right) = -R\omega^2\cos\theta\,\hat i - R\omega^2\sin\theta\,\hat j$$

The linear speed is $$v = \omega R \;\Rightarrow\; \omega^2 = \dfrac{v^2}{R^2}$$.

Substituting $$\omega^2$$ in the acceleration expression:

$$\vec a = -R\,\frac{v^2}{R^2}\cos\theta\,\hat i - R\,\frac{v^2}{R^2}\sin\theta\,\hat j = -\frac{v^2}{R}\cos\theta\,\hat i - \frac{v^2}{R}\sin\theta\,\hat j$$

Thus the centripetal acceleration always points toward the centre, opposite to the position vector.

Option C which is: $$-\dfrac{v^2}{R}\cos\theta\,\hat i - \dfrac{v^2}{R}\sin\theta\,\hat j$$