The equation of the normal to the curve $$y = (1+x)^{2y} + \cos^2(\sin^{-1}x)$$, at $$x = 0$$ is:

We have the implicit curve $$y=(1+x)^{2y}+\cos^{2}\!\bigl(\sin^{-1}x\bigr).$$

First we evaluate the point of contact by putting $$x=0.$$

The trigonometric term simplifies because $$\cos\!\bigl(\sin^{-1}x\bigr)=\sqrt{1-x^{2}},$$ so $$\cos^{2}\!\bigl(\sin^{-1}x\bigr)=1-x^{2}.$$

Therefore, at any $$x$$ we can write

$$y=(1+x)^{2y}+1-x^{2}.$$

Putting $$x=0$$ gives

$$y=(1+0)^{2y}+1-0^{2}=1^{2y}+1=1+1=2.$$

So the point on the curve is $$(0,2).$$

To find the slope of the tangent, we differentiate both sides with respect to $$x$$. We rewrite the power term by the exponential form so that a known formula can be applied:

$$(1+x)^{2y}=e^{2y\ln(1+x)}.$$

The derivative of $$e^{g(x)}$$ is $$e^{g(x)}\,g'(x).$$ Using this, we obtain

$$\frac{d}{dx}\bigl[(1+x)^{2y}\bigr]=(1+x)^{2y}\,\frac{d}{dx}\!\bigl[2y\ln(1+x)\bigr].$$

The derivative inside is found by the product rule:

$$\frac{d}{dx}\!\bigl[2y\ln(1+x)\bigr]=2\,\frac{dy}{dx}\,\ln(1+x)+\frac{2y}{1+x}.$$

The derivative of the right-hand side of the implicit equation is therefore

$$\frac{dy}{dx}=(1+x)^{2y}\Bigl[2\,\frac{dy}{dx}\,\ln(1+x)+\frac{2y}{1+x}\Bigr]-2x.$$

Now we substitute $$x=0$$ and $$y=2.$$ We note that $$\ln(1+0)=0,\quad(1+0)^{2y}=1,\quad\text{and}\quad-2x=-0.$$ Hence

$$\frac{dy}{dx}=1\Bigl[2\,\frac{dy}{dx}\,(0)+\frac{2\cdot2}{1}\Bigr]-0=4.$$

Thus the slope of the tangent at $$(0,2)$$ is $$m_t=4.$$ The slope of the normal is the negative reciprocal:

$$m_n=-\frac{1}{4}.$$

Using the point-slope form of a line, $$y-y_1=m(x-x_1),$$ with $$(x_1,y_1)=(0,2)$$ and $$m=m_n=-\frac{1}{4},$$ we get

$$y-2=-\frac{1}{4}(x-0).$$

Multiplying by $$4$$ to clear the fraction gives

$$4y-8=-x,$$

which rearranges to

$$x+4y=8.$$

Hence, the correct answer is Option C.