The function $$f(x)$$, that satisfies the condition $$f(x) = x + \int_0^{\pi/2} \sin x \cos y f(y) dy$$, is:

We start with the functional equation

$$f(x)=x+\int_{0}^{\pi/2}\sin x\,\cos y\,f(y)\,dy.$$

The variable of integration is $$y$$, so for the whole integral the factor $$\sin x$$ behaves like a constant. Hence

$$f(x)=x+\sin x\int_{0}^{\pi/2}\cos y\,f(y)\,dy.$$

This expression shows that $$f(x)$$ must be a sum of a term proportional to $$x$$ and a term proportional to $$\sin x$$. Guided by this observation, we assume

$$f(x)=x+A\sin x,$$

where $$A$$ is a constant to be determined. Now we substitute this trial form into the integral.

First we compute the integral

$$I=\int_{0}^{\pi/2}\cos y\,f(y)\,dy=\int_{0}^{\pi/2}\cos y\,(\,y+A\sin y\,)\,dy.$$

We split the integral into two simpler parts.

Part 1:

$$I_1=\int_{0}^{\pi/2}y\cos y\,dy.$$

Using integration by parts, with $$u=y$$ and $$dv=\cos y\,dy$$, we have $$du=dy$$ and $$v=\sin y$$. Therefore

$$I_1=y\sin y\Big|_{0}^{\pi/2}-\int_{0}^{\pi/2}\sin y\,dy

=\frac{\pi}{2}\cdot1-\bigl(-\cos y\bigr)\Big|_{0}^{\pi/2}

=\frac{\pi}{2}-\bigl[-\cos(\tfrac{\pi}{2})+ \cos(0)\bigr]

=\frac{\pi}{2}-\bigl[0-1\bigr]

=\frac{\pi}{2}-1.$$

Part 2:

$$I_2=A\int_{0}^{\pi/2}\cos y\,\sin y\,dy.$$

Using the substitution $$u=\sin y,\;du=\cos y\,dy$$, we get

$$I_2=A\int_{0}^{\pi/2}u\,du=A\left[\frac{u^{2}}{2}\right]_{0}^{\pi/2}

=A\left[\frac{\sin^{2}y}{2}\right]_{0}^{\pi/2}

=A\left[\frac{1^{2}}{2}-0\right]

=\frac{A}{2}.$$

Combining the two parts,

$$I=I_1+I_2=\left(\frac{\pi}{2}-1\right)+\frac{A}{2}.$$

Returning to the expression for $$f(x)$$, we substitute this value of $$I$$:

$$f(x)=x+\sin x\left(\frac{\pi}{2}-1+\frac{A}{2}\right).$$

But from our assumed form $$f(x)=x+A\sin x$$. Since the expression must hold for every $$x$$, the coefficients of $$\sin x$$ must be equal. Thus

$$A=\frac{\pi}{2}-1+\frac{A}{2}.$$

We now solve this simple linear equation for $$A$$. First bring like terms together:

$$A-\frac{A}{2}=\frac{\pi}{2}-1.$$

The left side simplifies to $$\frac{A}{2}$$, so

$$\frac{A}{2}=\frac{\pi}{2}-1.$$

Multiplying both sides by $$2$$ gives

$$A=\pi-2.$$

Finally, substituting $$A=\pi-2$$ into our trial form yields

$$f(x)=x+(\pi-2)\sin x.$$

This matches Option D in the list.

Hence, the correct answer is Option D.