The physical quantity that has the same dimensional formula as pressure is

Dimensional formula of pressure according to its definition,

$\left[\mathrm{P}\right]=\frac{\left[\text{Force}\right]}{\left[\text{Area}\right]}=\frac{{\mathrm{MLT}}^{-2}}{{\mathrm{L}}^2}={\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}$

Dimensional formula of force =[F]=[ma]= MLT^-2.

Dimensional formula of momentum = [mv]= MLT^-1.
Young’s modulus,

$\mathrm{Y}=\frac{\text{Stress}}{\text{Strain}}=\frac{\mathrm{F}/\mathrm{A}}{\mathrm{e}/\ell }$

$\text{where,}\frac{e}{\ell }=\frac{\text{elongation}}{\text{original length}}=\text{dimensionless.}$

$\text{So, }\left[\mathrm{Y}\right]=\frac{{\mathrm{MLT}}^{-2}}{{\mathrm{L}}^2}={\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}$

Coefficient of viscosity, according to the definition,  

$\begin{array}{rl} & F_v=-\eta A\frac{dV}{dx}\\ & \eta =\frac{F_{v}dx}{dVA}\end{array}$

where  F_V = viscous force 

$\left[\frac{dV}{dx}\right]=\left[\text{velocity gradiant}\right]=\frac{L{T}^{-1}}{L}=T^{-1}\text{.}$

A = area

$\text{So,}\left[\eta \right]=\frac{ML{T}^{-2}\mathrm{L}}{{\mathrm{LT}}^{-1}{\mathrm{L}}^2}={\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}$

So, correct option is that Young’s modulus of elasticity and pressure has the same dimensional formula.