Given that, 

$x=a\left(1+\cos \theta \right)$

$y=a\sin \theta$

Now,

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\left(\frac{dx}{d\theta }\right)}$

$\Rightarrow \frac{dy}{dx}=\frac{\frac{d}{d\theta }\left(a\sin \theta \right)}{\frac{d}{d\theta }\left(a\left(1+\cos \theta \right)\right)}$

$\Rightarrow \frac{dy}{dx}=\frac{a\cos \theta }{-a\sin \theta }$

$\Rightarrow \frac{dy}{dx}=-\cot \theta$

Equation of normal at $\left(x_1{y}_1\right)$ is

$\left(y-y_1\right)=-\frac{1}{\frac{dy}{dx}}\left(x-x_1\right)$

$\Rightarrow \left(y-a\sin \theta \right)=\frac{-1}{-\cot \theta }\left(x-\left(a\left(1+\cos \theta \right)\right)\right)$

$\Rightarrow \left(y-a\sin \theta \right)=\frac{x-a-\cos \theta }{\cot \theta }$

$\Rightarrow \frac{y-a\sin \theta }{x-a-a\cos \theta }=\frac{\sin \theta }{\cos \theta }$

$\Rightarrow \frac{y-a\sin \theta }{\sin \theta }=\frac{x-a-a\cos \theta }{\cos \theta }$

$\Rightarrow \frac{y}{\sin \theta }-\frac{a\sin \theta }{\sin \theta }=\frac{x}{\cos \theta }-\frac{a}{\cos \theta }-\frac{a\cos \theta }{\cos \theta }$

$\Rightarrow \frac{y}{\sin \theta }-a=\frac{x}{\cos \theta }-\frac{a}{\cos \theta }-a$

$\Rightarrow \frac{y}{\sin \theta }=\frac{x-a}{\cos \theta }$

Now by verifying options, $\left(a,0\right)$ satisfies the above condition

$\therefore$ The curve passes through $\left(a,0\right)$.