Step-by-step Solution:

1. Identify the Reactants and Products: The question states that white phosphorus ($P_4$) reacts with aqueous NaOH. One product is given as phosphine ($PH_3$). We need to identify the other phosphorus-containing compound and the reaction type. The reaction is a well-known disproportionation reaction of white phosphorus in a hot, concentrated alkali solution. The unbalanced chemical equation is: $P_4 + NaOH + H_2O \rightarrow PH_3 + NaH_2PO_2$ The other product is sodium hypophosphite ($NaH_2PO_2$).

2. Determine the Oxidation States: To classify the reaction and find the required oxidation states, we calculate the oxidation state (O.S.) of phosphorus in the reactant and products.

   • Reactant (White Phosphorus, $P_4$): In its elemental form, the oxidation state of phosphorus is 0.
   • Product (Phosphine, $PH_3$): Let the oxidation state of P be $x$. Hydrogen has an oxidation state of +1 when bonded to a more electronegative non-metal like phosphorus. $x + 3(+1) = 0 \implies x = -3$ So, the O.S. of P in $PH_3$ is -3.
   • Other Product (Sodium Hypophosphite, $NaH_2PO_2$): Let the oxidation state of P be $y$. Sodium (Na) is an alkali metal, so its O.S. is +1. Oxygen (O) generally has an O.S. of -2. Hydrogen (H) has an O.S. of +1. $(+1) + 2(+1) + y + 2(-2) = 0$ $1 + 2 + y - 4 = 0$ $y - 1 = 0 \implies y = +1$ So, the O.S. of P in $NaH_2PO_2$ is +1.

3. Determine the Reaction Type: In this reaction, the oxidation state of phosphorus changes from 0 in the reactant ($P_4$) to -3 in one product ($PH_3$) and +1 in another product ($NaH_2PO_2$).

   • $P(0) \rightarrow P(-3)$ is a reduction (gain of electrons).
   • $P(0) \rightarrow P(+1)$ is an oxidation (loss of electrons). Since the same element (phosphorus) from a single reactant is simultaneously oxidized and reduced, the reaction is a disproportionation reaction. A disproportionation reaction is a specific type of redox reaction.

4. Match with the Options: We are looking for the option that correctly identifies:

   • Reaction type: disproportionation reaction
   • O.S. of P in phosphine: -3
   • O.S. of P in the other product: +1

   Let's evaluate the given options:

   • A: redox reaction; $-$ 3 and $-$ 5. (Incorrect type and O.S.)
   • B: redox reaction; +3 and +5. (Incorrect type and O.S.)
   • C: disproportionation reaction; $-$ 3 and +1. (This matches our findings.)
   • D: disproportionation reaction; $-$ 3 and +3. (Incorrect O.S. for the other product.)

5. Conclusion: The reaction is a disproportionation reaction, and the oxidation states of phosphorus in phosphine and sodium hypophosphite are -3 and +1, respectively. Therefore, option C is the correct answer.

   The balanced chemical equation for the reaction is: $P_4(s) + 3 NaOH(aq) + 3 H_2O(l) \rightarrow PH_3(g) + 3 NaH_2PO_2(aq)$