Does the projectively stable category have projective modules? I am reading the book Elements of the Representation Theory of Associative Algebras: Volume 1 . On page 109, the projectively stable category is defined by $$ \underline{mod} A = mod A/\mathcal{P}. $$ The objects of $\underline{mod} A$ is the same as the objects in $mod A$ but the $K$-vector space $\underline{Hom}_A(M, N)$ of morphisms from $M$ to $N$ in $\underline{mod} A$ is defined to be the quotient vector space 
$$
\underline{Hom}_A(M, N) = Hom_A(M, N)/\mathcal{P}(M, N).
$$
Here $\mathcal{P}(M, N)$ is the subset of $Hom_A(M, N)$ consisting of all homomorphisms that factor through a projective $A$-module. 
By the definition, we have projective modules in $\underline{mod}A$. But it is said that $\underline{mod}A$ does not have projective modules in the book Representation Theory of Artin Algebras.
I am confused.
It is said in Proposition 2.2 on page 110 of the book Elements of the Representation Theory of Associative Algebras: Volume 1  that $M \mapsto Tr M$ induces a $K$-linear duality functor $Tr: \underline{mod}A \to \underline{mod}A^{op}$. If $\underline{mod}A$ has a projective module $P$, then $Tr P = 0$ and $Tr (Tr P) = Tr(0) = 0 \neq P$. This contradicts the fact that $Tr$ is a duality functor. So I think that $\underline{mod}A$ does not have a projective module. Is this true? Thank you very much.
 A: As I already mentioned in the comment, the projective objects get isomorphic to $0$ since the identity map $P\to P$ obviously factors through a projective module and therefore the identity map is the zero map, which is only true for the zero module. 
As you can see from that since every projective module gets "killed" when going to the stable module category you can leave them out before, i.e. you can construct the stable module category as a factor category of the full subcategory of all the non-projective modules. In that sense the stable module category does not contain projective modules.
If I am not mistaken this doesn't mean that the stable module category doesn't have projective objects, e.g. take $A=k[x]/x^2$. Then, up to isomorphism, in the stable module category of $A$ there is only one indecomposable object up to isomorphism (since the indecomposable projective gets isomorphic to $0$) and hence this should be a projective object (in the categorical sense).
For your second question: We have $\operatorname{Tr}\operatorname{Tr}(P)=0$ is correct, but $P\cong 0$, so this is no contradiction to being a duality.
