$\lim _{x \rightarrow 0} \frac{63^x-9^x-7^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=\ldots$.

To evaluate $\lim_{x \to 0} \frac{63^x - 9^x - 7^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}$, we first observe that substituting $x = 0$ gives the indeterminate form $\frac{0}{0}$ as both numerator and denominator vanish.

The numerator factors as $(9^x - 1)(7^x - 1)$ by writing $63^x$ as $9^x \cdot 7^x$ and grouping terms.

Rationalizing the denominator gives $\frac{1 - \cos x}{\sqrt{2} + \sqrt{1 + \cos x}}$, which using the identity $1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$ becomes $\frac{2 \sin^2 \left(\frac{x}{2}\right)}{\sqrt{2} + \sqrt{1 + \cos x}}$.

Thus, the limit becomes $\lim_{x \to 0} \frac{(9^x - 1)(7^x - 1)(\sqrt{2} + \sqrt{1 + \cos x})}{2 \sin^2 \left(\frac{x}{2}\right)}$.

Rewriting for standard limits:

$\lim_{x \to 0} \frac{9^x - 1}{x} \cdot \frac{7^x - 1}{x} \cdot \frac{x^2}{2 \sin^2 \left(\frac{x}{2}\right)} \cdot (\sqrt{2} + \sqrt{1 + \cos x})$

The standard limit $\lim_{x \to 0} \frac{a^x - 1}{x} = \log a$ gives $\log 9$ and $\log 7$ for the first two factors.
The third factor evaluates to $\lim_{x \to 0} \frac{x^2}{2 \sin^2 \left(\frac{x}{2}\right)} = 2$ using $\lim_{u \to 0} \frac{\sin u}{u} = 1$.
The fourth factor approaches $2\sqrt{2}$ as $x \to 0$.

Multiplying these results yields $(\log 9)(\log 7)(2)(2\sqrt{2}) = 4\sqrt{2} \log 7 \cdot \log 9$, which corresponds to option B.

Final answer: $\boxed{\text{B}}$