If $(P-<!--\; -\; -->\frac{a}{V^2})(V-<!--\; -\; -->b)=nRT$ , where P, V, R and T are pressure, volume, universal gas constant and temperature, then $\frac{a}{b^2}$ has same dimensional

The correct option is P.

To find the dimensions of the expression $\frac{a}{b^2}$, we use the principle of homogeneity of dimensions. According to this principle, physical quantities can be added or subtracted only if they have the same dimensions.

Let's analyze the terms inside the parentheses in the given equation:
$(P-\frac{a}{V^2}\left)\right(V-b)=nRT$

From the term $(V-b)$, the dimension of $b$ must be the same as the dimension of volume $V$:
$\left[b\right]=\left[V\right]$

From the term $(P-\frac{a}{V^2})$, the dimension of $\frac{a}{V^2}$ must be the same as the dimension of pressure $P$:
$\left[\frac{a}{V^2}\right]=\left[P\right]$
This gives the dimension of $a$ as:
$\left[a\right]=\left[P\right]·\left[V^2\right]$

Now, we substitute the dimensions of $a$ and $b$ into the target expression $\frac{a}{b^2}$:
$\left[\frac{a}{b^2}\right]=\frac{\left[P\right]·\left[V^2\right]}{{\left[b\right]}^2}$

Since $\left[b\right]=\left[V\right]$, we have ${\left[b\right]}^2=\left[V^2\right]$:
$\left[\frac{a}{b^2}\right]=\frac{\left[P\right]·\left[V^2\right]}{\left[V^2\right]}=\left[P\right]$

Therefore, the expression $\frac{a}{b^2}$ has the same dimension as pressure, which is represented by P.