1. Interpret the expression

We need to find

$$\log_e 2\; \frac{d}{dx}\left(\log_{\cos x}(\cosec x)\right)$$

at $x=\frac{\pi}{4}$.

Since $\log_e 2 = \ln 2$, the quantity is

$$(\ln 2)\, \frac{d}{dx}\left(\log_{\cos x}(\cosec x)\right).$$

2. Convert logarithm to natural logarithm

Using

$$\log_a b = \frac{\ln b}{\ln a},$$

we get

$$\log_{\cos x}(\cosec x)=\frac{\ln(\cosec x)}{\ln(\cos x)}.$$

Now simplify each term:

$$\ln(\cosec x)=\ln\left(\frac{1}{\sin x}\right)=-\ln(\sin x),$$ $$\ln(\cos x)=\ln(\cos x).$$

So,

$$\log_{\cos x}(\cosec x)=\frac{-\ln(\sin x)}{\ln(\cos x)}.$$

At $x=\frac{\pi}{4}$, since $\sin x=\cos x$, this expression becomes

$$\frac{-\ln(\sin x)}{\ln(\sin x)}=-1.$$

But we need its derivative.

3. Differentiate using quotient rule

Let

$$f(x)=\frac{\ln(\cosec x)}{\ln(\cos x)}.$$

Take

$$u=\ln(\cosec x), \qquad v=\ln(\cos x).$$

Then

$$u' = \frac{d}{dx}[\ln(\cosec x)] = -\cot x,$$

because

$$\frac{d}{dx}(\ln \cosec x)=\frac{1}{\cosec x}(-\cosec x\cot x)=-\cot x.$$

Also,

$$v' = \frac{d}{dx}[\ln(\cos x)] = -\tan x.$$

Now by quotient rule,

$$f'(x)=\frac{\nu'v-\nu v'}{v^2} =\frac{(-\cot x)\ln(\cos x)-\ln(\cosec x)(-\tan x)}{[\ln(\cos x)]^2}.$$

So,

$$f'(x)=\frac{-\cot x\,\ln(\cos x)+\tan x\,\ln(\cosec x)}{[\ln(\cos x)]^2}.$$

4. Evaluate at $x=\frac{\pi}{4}$

At $x=\frac{\pi}{4}$,

$$\tan\frac{\pi}{4}=1, \qquad \cot\frac{\pi}{4}=1,$$ $$\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}, \qquad \cosec\frac{\pi}{4}=\sqrt{2}.$$

Thus,

$$\ln\left(\cos\frac{\pi}{4}\right)=\ln\left(\frac{1}{\sqrt{2}}\right)=-\frac12\ln 2,$$ $$\ln\left(\cosec\frac{\pi}{4}\right)=\ln(\sqrt{2})=\frac12\ln 2.$$

Substitute into $f'(x)$:

$$f'\left(\frac{\pi}{4}\right) =\frac{-1\cdot\left(-\frac12\ln 2\right)+1\cdot\left(\frac12\ln 2\right)}{\left(-\frac12\ln 2\right)^2}.$$

The numerator is

$$\frac12\ln 2+\frac12\ln 2=\ln 2.$$

The denominator is

$$\left(\frac12\ln 2\right)^2=\frac14(\ln 2)^2.$$

Hence,

$$f'\left(\frac{\pi}{4}\right)=\frac{\ln 2}{\frac14(\ln 2)^2}=\frac{4}{\ln 2}.$$

5. Multiply by $\ln 2$

The required value is

$$(\ln 2)\, f'\left(\frac{\pi}{4}\right) =(\ln 2)\cdot \frac{4}{\ln 2}=4.$$

6. Match with options

Thus the correct option is:

$$\boxed{4}$$

which is Option D.