7. Given the parabola $C: y^{2}=8 x$ with focus $F$, the directrix intersects the $x$-axis at point $K$. Point $A$ is on the parabola $C$, and $|A K|=\sqrt{2}|A F|$. Then the area of $\triangle A F K$ is ( ).
(A) 4
(B) 8
(C) 16
(D) 32
7. B.

Since the focus of the parabola $C: y^{2}=8 x$ is $F(2,0)$, and the directrix is $x=-2$, we have $K(-2,0)$.

Let $A\left(x_{0}, y_{0}\right)$. Draw a perpendicular line $A B$ from point $A$ to the directrix, then $B\left(-2, y_{0}\right)$.
Given that $|A K|=\sqrt{2}|A F|$, and
$$
A F=A B=x_{0}-(-2)=x_{0}+2 \text {, }
$$

thus, from $B K^{2}=A K^{2}-A B^{2}$, we get
$$
y_{0}^{2}=\left(x_{0}+2\right)^{2} \text {, }
$$

which means $8 x_{0}=\left(x_{0}+2\right)^{2}$.
Solving this, we get $A(2, \pm 4)$.
Therefore, $S_{\triangle A F K}=\frac{1}{2}|K F|\left|y_{0}\right|=\frac{1}{2} \times 4 \times 4=8$.