To solve this problem, we need to determine which of the given options is a subset of the power set of $A$, denoted as $\mathcal{P}(A)$.

Step 1: Determine the elements of $A$.

Given: $A = \{1, 2, \{1\}\}$

Step 2: Find the power set $\mathcal{P}(A)$.

The power set $\mathcal{P}(A)$ is the set of all subsets of $A$. These subsets include:

1. The empty set $\emptyset$
2. The singleton sets $\{1\}$, $\{2\}$, and $\{\{1\}\}$
3. The pairs $\{1, 2\}$, $\{1, \{1\}\}$, and $\{2, \{1\}\}$
4. The entire set $A = \{1, 2, \{1\}\}$

Thus: $\mathcal{P}(A) = \{\emptyset, \{1\}, \{2\}, \{\{1\}\}, \{1, 2\}, \{1, \{1\}\}, \{2, \{1\}\}, \{1, 2, \{1\}\}\}$

Step 3: Analyze the options.

Now, we check if any of the provided options is a subset of $\mathcal{P}(A)$.

• Option a: $\{1\}$
  $\{1\}$ is indeed an element of $\mathcal{P}(A)$, but $\{1\}$ as a set is not a subset of $\mathcal{P}(A)$, it's an element.

• Option b: $\{1, 2, \{1, 2\}\}$
  This set contains $1$, $2$, and $\{1, 2\}$. $\{1, 2\}$ is not an element of $\mathcal{P}(A)$, so this option cannot be a subset.

• Option c: $\{\{\{1\}\}\}$
  This set contains $\{\{1\}\}$, but $\{\{1\}\}$ is not an element of $\mathcal{P}(A)$, so this option cannot be a subset.

• Option d: $\{\{\{2\}\}\}$
  This set contains $\{\{2\}\}$. $\{\{2\}\}$ is not an element of $\mathcal{P}(A)$, so this option cannot be a subset.

Conclusion:

None of the options presented are subsets of $\mathcal{P}(A)$. But among the options, $\{1\}$ (option a) is an element of $\mathcal{P}(A)$. However, none of the options fit the definition of being a subset of $\mathcal{P}(A)$ if we're strictly checking for a subset.

If I had to select the most correct answer given the choices, it would be Option a, but strictly speaking, this answer is based on element inclusion, not a subset.