Which of the following function(s) not defined at x=0 has/have irremovable discontinuity at x=0?

The function is not defined at $x=0$ because $\ln|0|$ is undefined.
Let's evaluate the limit as $x \to 0$:
As $x \to 0$, $|x| \to 0^+$.
As $y \to 0^+$, $\ln y \to -\infty$.
So, $\lim_{x \to 0} \ln|x| = -\infty$.
Therefore, $\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1}{\ln|x|} = \frac{1}{-\infty} = 0$.
Since the limit exists but the function is not defined at $x=0$, this is a removable discontinuity.

The function is not defined at $x=0$ because the denominator $x$ is zero.
Let's evaluate the limit as $x \to 0$:
Consider the inner function $g(x) = \frac{|\sin x|}{x}$.
For $x > 0$ (as $x \to 0^+$), $\sin x > 0$, so $|\sin x| = \sin x$.
$\lim_{x \to 0^+} \frac{\sin x}{x} = 1$.
For $x < 0$ (as $x \to 0^-$), $\sin x < 0$, so $|\sin x| = -\sin x$.
$\lim_{x \to 0^-} \frac{-\sin x}{x} = -1$.
Now, evaluate the limit of $f(x)$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \cos\left(\frac{|\sin x|}{x}\right) = \cos(1)$.
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \cos\left(\frac{|\sin x|}{x}\right) = \cos(-1) = \cos(1)$.
Since the left-hand limit and right-hand limit are equal, $\lim_{x \to 0} f(x) = \cos(1)$.
Since the limit exists but the function is not defined at $x=0$, this is a removable discontinuity.

The function is not defined at $x=0$ because $\frac{\pi}{x}$ is undefined.
Let's evaluate the limit as $x \to 0$:
We know that $-1 \le \sin\left(\frac{\pi}{x}\right) \le 1$.
Multiplying by $x$:
If $x > 0$, then $-x \le x \sin\left(\frac{\pi}{x}\right) \le x$.
If $x < 0$, then $x \le x \sin\left(\frac{\pi}{x}\right) \le -x$.
In both cases, $|x \sin\left(\frac{\pi}{x}\right)| \le |x|$.
As $x \to 0$, $|x| \to 0$. By the Squeeze Theorem, $\lim_{x \to 0} x \sin\left(\frac{\pi}{x}\right) = 0$.
Since the limit exists but the function is not defined at $x=0$, this is a removable discontinuity.

The function is not defined at $x=0$ because $\cot x = \frac{\cos x}{\sin x}$ is undefined at $x=0$.
Let's evaluate the limit as $x \to 0$:
Consider the behavior of $\cot x$ as $x \to 0$:
As $x \to 0^+$, $\sin x \to 0^+$ and $\cos x \to 1$, so $\cot x \to +\infty$.
As $x \to 0^-$, $\sin x \to 0^-$ and $\cos x \to 1$, so $\cot x \to -\infty$.
Now, evaluate the limits of $f(x)$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{1+2^{\cot x}}$. As $\cot x \to +\infty$, $2^{\cot x} \to +\infty$.
So, $\lim_{x \to 0^+} f(x) = \frac{1}{1+\infty} = 0$.
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{1+2^{\cot x}}$. As $\cot x \to -\infty$, $2^{\cot x} \to 0$.
So, $\lim_{x \to 0^-} f(x) = \frac{1}{1+0} = 1$.
Since the left-hand limit (1) and the right-hand limit (0) are different, the limit $\lim_{x \to 0} f(x)$ does not exist. This indicates an irremovable (jump) discontinuity.

All four functions are not defined at $x=0$.
Functions (A), (B), and (C) have removable discontinuities at $x=0$ because their limits exist.
Function (D) has an irremovable (jump) discontinuity at $x=0$ because its left-hand and right-hand limits are different.

The question asks for the function(s) not defined at $x=0$ that has/have an irremovable discontinuity at $x=0$. Only function (D) fits this description.

The final answer is $\text{D}$.