Let $\Gamma$ denote the modular group $PSL(2,\mathbb{Z})$.
We define $\Gamma(N)$ to be the kernel of following natural map $$PSL(2,\mathbb{Z})\rightarrow PSL(2,\mathbb{Z}/N\mathbb{Z})$$
Now, $$\Gamma(1)=PSL(2,\mathbb{Z})$$
Let $X(N)$ demote the quotient of complex upper half plane under the action of $\Gamma(1)$.
Is it correct that $X(1) $ is isomorphic $P^{1}$ ?
If yes, kindly mention the reference.
The $j$-invariant is a $\Gamma$ invariant holomorphic and surjective map from $\mathbb H\to\mathbb C$, and descends to a holomorphic bijection $$j:X(1)\to\mathbb C$$
which has a pole at infinity.
As you've defined $X(1)$ (which is usually denoted $Y(1)$), it is clear that it is not isomorphic to $\mathbb P^1(\mathbb C)$ for the simple reason that it is not compact - it has a "cusp" at infinity.
However, $\Gamma(N)$ acts on $\mathbb H^* = \mathbb H\cup\mathbb P^1(\mathbb Q)$. Letting $$X(N) :=\Gamma(N)\backslash \mathbb H^*,$$$X(N)$ is now compact, and the $j$-invariant becomes a holomorphic bijection $X(1)\to\mathbb P^1(\mathbb C)$ with holomorphic inverse. In particular, $X(1)\cong\mathbb P^1(\mathbb C)$ as Riemann surfaces.