Analysis of each statement:

Statement A: Upon heating, $H_3PO_3$ undergoes disproportionation reaction to produce $H_3PO_4$ and $PH_3$.

1. Identify Oxidation States:

   • In phosphorous acid ($H_3PO_3$), let the oxidation state of P be $x$. We have $3(+1) + x + 3(-2) = 0$, which gives $x = +3$.
   • In phosphoric acid ($H_3PO_4$), let the oxidation state of P be $y$. We have $3(+1) + y + 4(-2) = 0$, which gives $y = +5$.
   • In phosphine ($PH_3$), let the oxidation state of P be $z$. We have $z + 3(+1) = 0$, which gives $z = -3$.

2. Analyze the Reaction: The reaction is the thermal decomposition of $H_3PO_3$: $4H_3PO_3 \xrightarrow{\Delta} 3H_3PO_4 + PH_3$. In this reaction, the oxidation state of phosphorus changes from +3 in $H_3PO_3$ to +5 in $H_3PO_4$ (oxidation) and to -3 in $PH_3$ (reduction). A reaction where the same element is both oxidized and reduced is called a disproportionation reaction.

3. Conclusion: The statement is correct.

Statement B: While $H_3PO_3$ can act as reducing agent, $H_3PO_4$ cannot.

1. Analyze $H_3PO_3$: The oxidation state of P in $H_3PO_3$ is +3. This is an intermediate oxidation state for phosphorus, which can range from -3 to +5. Since phosphorus can be oxidized from +3 to a higher state (like +5), $H_3PO_3$ can act as a reducing agent. For example, it reduces silver nitrate to silver: $H_3PO_3 + 2AgNO_3 + H_2O \rightarrow H_3PO_4 + 2Ag + 2HNO_3$.

2. Analyze $H_3PO_4$: The oxidation state of P in $H_3PO_4$ is +5. This is the highest possible oxidation state for phosphorus. Since it cannot be further oxidized, $H_3PO_4$ cannot act as a reducing agent.

3. Conclusion: The statement is correct.

Statement C: $H_3PO_3$ is a monobasic acid.

1. Determine Basicity from Structure: The basicity of an oxoacid is determined by the number of ionizable hydrogen atoms, which are those attached to oxygen atoms (forming -OH groups). The structure of phosphorous acid ($H_3PO_3$) is:

   $$\begin{array}{c} \text{O} \\ \|\| \text{H} - \text{P} - \text{OH} \\ | \\ \text{OH} \end{array}$$

   The structure has two P-OH bonds and one P-H bond. Only the two hydrogens in the P-OH groups are acidic and ionizable.

2. Conclusion: Since there are two ionizable hydrogen atoms, $H_3PO_3$ is a dibasic acid, not monobasic. Therefore, the statement is incorrect.

Statement D: The H atom of P-H bond in $H_3PO_3$ is not ionizable in water.

1. Analyze Bond Polarity: As seen in the structure for statement C, $H_3PO_3$ has one P-H bond. The electronegativity of phosphorus (2.19) and hydrogen (2.20) are very similar. This results in a nearly non-polar covalent bond. The hydrogen atom is not released as a proton ($H^+$) in water because the bond is not sufficiently polarized.

2. Compare with O-H bond: In contrast, the O-H bonds are highly polar due to the large electronegativity difference between oxygen (3.44) and hydrogen (2.20), making these hydrogens acidic and ionizable.

3. Conclusion: The statement is correct. The presence of a non-ionizable P-H bond is also what imparts reducing character to $H_3PO_3$.

Final Summary: Statements A, B, and D are correct, while statement C is incorrect.