Suppose $\mathcal{V}$ be a subspaces of $\mathbb{R}^n$, Suppose $\mathcal{V}$ be a subspaces of $\mathbb{R}^n$, 
Suppose $P: \mathbb{R}^n\to\mathbb{R}^n$ be a projection, then I need to prove the following
$$ \mathcal{V}\cap\text {im } P=P^{-1}\mathcal{V}\cap\text{im }P$$
Suppose $x\in \mathcal{V}\cap\text {im } P\Rightarrow x\in\mathcal{V}\text { and } \exists y\in \text{ im } P\ni x=Py\in \mathcal{V}\Rightarrow Px=Py=x=P^{-1}x\in P^{-1}\mathcal{V}\Rightarrow x\in P^{-1}\mathcal{V}\cap im P$
now, suppose $x\in P^{-1}\mathcal{V}\cap im P$ so $Py=x$
I am not able to prove please help.

I would go and try to prove both inclusions separately. 
So let's focus on $$ (\mathcal{V}\cap\text {im } P) \subset (P^{-1}\mathcal{V}\cap\text{im }P) $$
first. 
$$ x \in (\mathcal{V}\cap\text {im } P) \Rightarrow x \in V \land x \in {im } P $$ 
 $$ \Rightarrow   x \in V \land \exists y \in \mathbb{R}^n :  Py=x  $$
 By the definition of a projection we get: 
$$ \Rightarrow   x \in V \land Py = Px = x $$
$$ \Rightarrow  x \in (P^{-1}\mathcal{V}\cap\text{im }P) $$
Now for the second part I'll give you the following hint because P is a projection we know:
 $$ x \in \text {im } P \iff Px = x $$ 
I hope that will help you a bit. 
