1. Condition for geometrical isomerism
   A compound shows geometrical ($E/Z$ or cis-trans) isomerism about a double bond only if each carbon of the $C=C$ bond has two different substituents.

2. Check each option

Option A: 3-Phenyl-1-butene

Structure: $\mathrm{CH_2=CH-CH(Ph)-CH_3}$ For the double bond:

• Left carbon = $\mathrm{CH_2}$, substituents are $H$ and $H$
• Since one double-bond carbon has two identical substituents, geometrical isomerism is not possible.

Option B: 2-Phenyl-1-butene

Structure: $\mathrm{CH_2=C(Ph)-CH_2-CH_3}$ For the double bond:

• Left carbon = $\mathrm{CH_2}$, substituents are $H$ and $H$
• Again, one double-bond carbon has identical substituents, so geometrical isomerism is not possible.

Option C: 1,1-Diphenyl-1-propane

Structure: $\mathrm{(Ph)_2C-CH_2-CH_3}$ This is a saturated compound; there is no $C=C$ double bond to restrict rotation. Hence geometrical isomerism is not possible.

Option D: 1-Phenyl-2-butene

Structure: $\mathrm{Ph-CH_2-CH=CH-CH_3}$ For the double bond:

• Left double-bond carbon has substituents: $H$ and $\mathrm{CH_2Ph}$
• Right double-bond carbon has substituents: $H$ and $\mathrm{CH_3}$

Each carbon of the double bond has two different groups, so geometrical isomerism is possible.

3. Conclusion Only Option D exhibits geometrical isomerism.