Step 1: Identify the nature of the reagents.
Concentrated $HNO_3$ is a powerful oxidising agent. So when phosphorus meets it, phosphorus is oxidised while nitrogen of $HNO_3$ is reduced.
Step 2: Track the oxidation of phosphorus.
Elemental phosphorus starts at oxidation state $0$. On oxidation by hot concentrated $HNO_3$ it reaches its highest common acid, phosphoric acid $H_3PO_4$, where phosphorus is in the $+5$ state.
Step 3: Confirm the oxidation number in $H_3PO_4$.
Taking hydrogen as $+1$ and oxygen as $-2$, for $H_3PO_4$ we get $3(+1) + x + 4(-2) = 0$, giving $x = +5$. This confirms phosphorus is oxidised from $0$ to $+5$.
Step 4: Track the reduction of nitrogen.
In concentrated $HNO_3$ nitrogen is at $+5$. The concentrated acid is typically reduced only to $NO_2$, where nitrogen is $+4$.
Step 5: Confirm the oxidation number in $NO_2$.
For $NO_2$, $x + 2(-2) = 0$ gives $x = +4$, so nitrogen drops from $+5$ to $+4$, which is the reduced product.
Step 6: State the oxidised and reduced products.
The oxidised product is $H_3PO_4$ and the reduced product is $NO_2$. So the pair is
\[ \boxed{H_3PO_4,\; NO_2} \]