If the Boolean expression $(\mathrm{p}\wedge \mathrm{q})⊛(\mathrm{p}\otimes \mathrm{q})$ is a tautology, then $⊛$ and $\otimes$ are respectively given by

Option $\left(A\right)$

$(p\wedge q)\to (p\to q)$

We know that $a\to b=~a\vee b$

$=(p\wedge q)⟶(~p\vee q)$

$=~(p\wedge q)\vee (~p\vee q)$

Using De Morgan's Law i.e. $~\left(a\wedge b\right)=~a\vee ~b,$

$=(~p\vee ~q)\vee (~p\vee q)$

Using associative law and commutative law i.e. $\left(a\vee b\right)\vee c=a\vee \left(b\vee c\right) \& a\vee b=b\vee a,$

$=~p\vee (~q\vee q)$

$=~p\vee t$

$=t$

Option $\left(B\right)$

$(p\wedge q)\wedge (p\vee q)=(p\wedge q)$

Which is not a tautology.

Option $\left(C\right)$

$(p\wedge q)\vee (p\to q)$

$=(p\wedge q)\vee (~p\vee q)$

$=~p\vee q$

Which is not a tautology.

Option $\left(D\right)$

$(p\wedge q)\wedge (p\to q)$

$=(p\wedge q)\wedge (~p\vee q)$

$=p\wedge q$

Which is not a tautology.

Hence, the correct answer is option $\left(A\right).$