On the kuratowski definition of ordered pairs. Kuratowski gave the following definition for ordered pair,
$$(a,b)=\{\{a\},\{a,b\}\}$$
Coming to it's geometric explanation, I think of it as a point in Cartesian plane.
Now similarly, what if we take only one argument inside my ordered pair?
My argument is that,
$$(a)=a$$
Because intuitively, $(a)$ is a point on 1-dimensional plane i.e number line.
And at the same time can we use kuratowski definition to prove it(if my argument is true)?
 A: Define an "assignment" (i.e. a class function) $F$ by
$$F(x,y)=\{\{x\},\{x,y\}\}$$
You can think of $F(x,y)$ as a set-theoretical encoding from the inputs $x$ and $y$. The encoding has the desired meaning of an "ordered pair" since there are two projection functions $\pi_1$ and $\pi_2$ that satisfy the following property: $\pi_1(F(x,y))=x$ and $\pi_2(F(x,y))=y$. You can think of $\pi_1$ and $\pi_2$ as set-theoretical decoders that allow us to identify what the first coordinate is (by applying $\pi_1$) or to identify what the second coordinate is (by applying $\pi_2$). I forgo defining $\pi_1$ and $\pi_2$ explicitly here.
Now you want to introduce another "assignment" $G$ defined by
$$G(x)=\boxed{???}$$
I put three question marks here because you didn't actually define it, but the important thing I gathered was that it takes one input. The map $F$ requires two inputs therefore $F\ne G$. Then you want to argue that $G(x)=x$ and you believe Kuratowski's $F$ map is somehow relevant. I hope I made it clear why your question is perceived as nonsensical.
I'll play along and assume you want to propose $G(x)=x$ by definition. You can think of $G(x)$ as the trivial encoding where nothing happens. Then the projection function $\pi_1$ can also do nothing, i.e., $\pi_1(y)=y$. Then indeed $\pi_1(G(x))=x$ since both $ G $ and $\pi_1$ do nothing to the input $x$.
