`X overset(Sn//HCl) to overset(NaNO_(2))underset(HCl)to Z overset(CuCN)underset(Delta)to (P) overset(H_(3)O)to Ph-COOH` correct options is/are:

Options

A. X is Ph`-NH_(2)`
B. Y is `Ph-NO_(2)`
C. Z is `Ph-NH_(2)`
D. P is Ph-CN

Explanation

To solve the problem step by step, we will analyze the transformations occurring from compound X to the final product Ph-COOH.

### Step 1: Identify Compound X
The first reagent is Sn/HCl, which is a reducing agent. It is used to reduce nitro groups (NO2) to amine groups (NH2). Therefore, we can conclude that:
- **X = Ph-NO2 (Nitrobenzene)**

### Step 2: Reduction of X to Y
When X (Ph-NO2) is treated with Sn/HCl, it is reduced to an amine:
- **Y = Ph-NH2 (Aniline)**

### Step 3: Diazotization of Y to Z
Next, Y is treated with NaNO2 in HCl. This reaction is known as diazotization, where an amine is converted to a diazonium salt:
- **Z = Ph-N2^+ Cl^- (Benzene diazonium chloride)**

### Step 4: Reaction of Z with CuCN to form P
The diazonium salt Z is then treated with CuCN, which introduces a cyanide group (CN) to the aromatic ring:
- **P = Ph-C≡N (Benzene cyanide)**

### Step 5: Hydrolysis of P to form the final product
The final step involves the hydrolysis of P (Ph-C≡N) using H3O^+. This reaction converts the nitrile group (C≡N) into a carboxylic acid (COOH):
- **Final Product = Ph-COOH (Benzoic acid)**

### Summary of the transformations:
- **X = Ph-NO2**
- **Y = Ph-NH2**
- **Z = Ph-N2^+ Cl^-**
- **P = Ph-C≡N**
- **Final Product = Ph-COOH**

### Correct Option
From the analysis, we can conclude that the only correct statement is:
- P is Ph-C≡N, which is correct.