<s>[INST] Is this the only solution to the problem? Find all mutually non-equivalent $A(p,q)$ so that $(Aq\rightarrow \overline{p}) = (p+A)$ is true regardless of $p$ and $q$, where $\overline{x}$ is negation, $xy$ is conjunction, $x+y$ disjunction, $x\rightarrow y$ implication and $x=y$ equivalence.
\begin{matrix}p&q&A&(Aq\rightarrow \overline{p}) = (p+A)\\
0&0&0&0\\
0&0&1&1\\
0&1&0&0\\
0&1&1&1\\
1&0&0&1\\
1&0&1&1\\
1&1&0&1\\
1&1&1&0
\end{matrix}
From the truth table we see that: $((Aq\rightarrow \overline{p}) = (p+A)) = \overline{p}\overline{q}A + \overline{p}qA + p\overline{q}\overline{A} + p\overline{q}A + pq\overline{A}$
$$\overline{p}\overline{q}A + \overline{p}qA + p\overline{q} + pq\overline{A} = 1$$
\begin{align}
A&=\overline{p}\overline{q}+\overline{p}q+\overline{(pq)}=\\
 &=\overline{p}+\overline{p}+\overline{q}=\\
 &=\overline{p}+\overline{q}
\end{align}
Cheking result
$$((\overline{p}+\overline{q})q\rightarrow \overline{p})=p+(\overline{p}+\overline{q})$$
For $p=0$ and $q=0$
$$
((1+1)0\rightarrow 1)=0+(1+1)\\
(0\rightarrow 1) = 1 \\
1=1 \\
$$
For $p=0$, $q=1$
$$
((1+0)1\rightarrow 1)=0+(1+0)\\
(1\rightarrow 1)=1 \\
1=1 \\
$$
For $p=1$, $q=0$
$$
((0+1)0\rightarrow 0)=1+(0+1)\\
(0\rightarrow 0)=1 \\
1=1 \\
$$
For $p=1$, $q=1$
$$
((0+0)1\rightarrow 0)=1+(0+0)\\
(0\rightarrow 0)=1 \\
1=1 \\
$$
This is a correct solution, but is it the only one?
I ask because the question uses the word all, and though that may not mean anything, I'd just like to check in case I missed something.
 [/INST] It seems to me you are correct up to here:
$$\overline{p}\overline{q}A + \overline{p}qA + p\overline{q} + pq\overline{A}
 = 1.$$
It does not follow that
$A=\overline{p}\overline{q}+\overline{p}q+\overline{(pq)} 
= \overline p + \overline q,$ however.
Yes, it is true that
$$
(A = \overline p + \overline q)\rightarrow 
(\overline{p}\overline{q}A + \overline{p}qA + p\overline{q} + pq\overline{A}
 = 1)
$$
for all possible truth assignments,
but the converse is not true.
The truth table shows that $A$ is one of two possible non-equivalent expressions, because for just one assignment to $p$ and $q,$
namely, $p=1, q=0,$ you have two choices for the value of $A.$

In particular, try the substitution $A = \overline p.$
With that substitution,
$Aq\rightarrow \overline p$ becomes
$\overline pq\rightarrow \overline p$, a tautology,
and $p+A$ becomes
$p+\overline p$, also a tautology,
so $(Aq\rightarrow \overline p)=(p+A).$
You can find this solution (rather than having to guess it) by selecting rows $2$, $4$, $5$, and $7$ from the truth table.
</s>