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Is the correspondence \begin{align*}f: \mathcal{P}(X)&\to \mathcal{P}(\mathcal{P}(X))\\A&\mapsto \mathcal{P}(A)\end{align*} a function?

If it is, I am thinking of using it to solve this problem

Let $(A_i)_{i\in I}$ be a family of subsets of a set $X$. Are the relations: $$\mathcal{P}\big(\bigcap_{i\in I}A_i\big) = \bigcap_{i\in I}\mathcal{P}(A_i)\\\mathcal{P}\big(\bigcup_{i\in I}A_i\big) = \bigcup_{i\in I}\mathcal{P}(A_i)$$ true?

If $f$ is a function, then the answer is quick and easy (it would need to be injective for the first relation to be true, and it seems it would be).

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    It's possible that set theorists will take some issue with your definition of function, but I'd say you're already looking at it. – 2017-02-25

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Yes, it's a function. I don't see how it is useful for solving your problem. If I were faced with those questions, instead of trying to find some way to state them in more arcane and confusing language, I'd be trying to state them more simply and clearly. In plain language, the questions are:

Is it true that something is a subset of the intersection of a bunch of sets if and only if it's a subset of each of them?

Is it true that a set is a subset of the union of a bunch of sets if and only if it's a subset of one of them?

Offhand I'd guess that one and a half of those statements are true.

P.S. Although $\mathcal P$ is a function, your idea for answering the questions is wrong. First let me explain your notational confusion.

Consider a function $f:X\to Y.$ If $a$ is an element of $X,$ the symbol $f(a)$ denotes (as usual) the value of the function at $a,$ and is an element of $Y.$ On the other hand, if $A$ is a subset of $X,$ then $f[A]$ (note the square brackets) denotes the set $\{f(a):a\in A\}$ and is a subset of $Y.$

We often ignore this distinction and use round brackets $f(\cdot)$ in both cases, and usually this does no harm. However, it gets confusing if the elements of $X$ are themselves sets, as they are here. In particular, the identity $f[A\cup B]=f[A]\cup f[B]$ (note square brackets) is always true, but $f(A\cup B)=f(A)\cup f(B)$ is usually false.

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    I agree with a simple and straightforward approach is best. But just to clarify what I was trying to do, I wanted to use $\mathcal{P}\big(\bigcup_{i\in I}A_i\big) = f\big(\bigcup_{i\in I}A_i\big)=\bigcup_{i\in I}f(A_i) = \bigcup_{i\in I}\mathcal{P}(A_i)$ which works for any function $f$. For the intersection, we get an inclusion, and equality if $f$ is injective. – 2017-02-25
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    Oh my. You have managed to confuse yourself with fancy notation. To start with, the statement about unions is ***FALSE***.For example, suppose $I=\{1,2,3,4,5\}$ and let $A_1,A_2,A_3,A_4,A_5$ be disjoint one-element sets. Then $\bigcup_{i\in I}A_i$ is a $5$-element set, so $\mathcal P(\bigcup_{i\in I}A_i)$ had $32$ elements; but $\bigcup_{i\in I}\mathcal P(A_i)$ has only $6$ elements. – 2017-02-25
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    The common notation $$f(A)=\{f(a):a\in A\}$$ can lead to confusion because $f(\cdot)$ is used to mean two different things. Using a more precise notation such as $$f[A]=\{f(a):a\in A\}$$ it would be easier to explain what you're doing wrong: $$f[A\cup B]=f[A]\cup f[B]$$ is true always, but $$f(A\cup B)=f(A)\cup f(B)$$ hardly ever. – 2017-02-25
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    @bof: While not an established notation, I think that something like $f^\in(A)$ would be a clearer notation: We a re looking at a *different* function that is determined by the original; the function $f:X\to Y$ gives rise to the function $f^\in:\mathcal P(X)\to\mathcal P(Y), A\mapsto\{f(x):x\in A\}$. – 2017-02-25
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    @bof you are right, I did confuse myself. Can you clarify one thing: using your notations, for a set $A$, I understand $f[A]$, but what does $f(A)$ mean? – 2017-02-25
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    @celtschk thanks for your comment, it definitely helped understand what I did wrong. – 2017-02-25
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    In any case, I understand now how my approach was wrong. Even considering a function like $f^{\in}$ wouldn't help solve my problem. I'll just go at it the simple way. Thanks – 2017-02-25
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    If the domain of the function $f$ is a set of sets, and the set $A$ is an element of that domain, then naturally $f(A)$ is just the value of $f$ at $A.$ For example, $f(A)$ might be the power set of $A.$ – 2017-02-25