Choose an injective resolution $\underline{M} \to I^\bullet$ in $\mathcal{P}$, see Lemma \ref{lemma-enough-injectives}. By Derived Categories, Lemma \ref{derived-lemma-compute-ext-resolutions} any element of $\Ext^i_\mathcal{P}(\underline{P}, \underline{M})$ comes from a morphism $\varphi : \underline{P} \to I^i$ with $d^i \circ \varphi = 0$. We will prove that the Yoneda extension $$ E : 0 \to \underline{M} \to I^0 \to \ldots \to I^{i - 1} \times_{\Ker(d^i)} \underline{P} \to \underline{P} \to 0 $$ of $\underline{P}$ by $\underline{M}$ associated to $\varphi$ is trivial, which will prove the lemma by Derived Categories, Lemma \ref{derived-lemma-yoneda-extension}. \medskip\noindent For $F$ a module-valued functor on $\textit{Alg}_A$ say $(*)$ holds if for all $B \in \Ob(\textit{Alg}_A)$ the functor $TF(B, -)$ on $B$-modules transforms a short exact sequence of $B$-modules into a right exact sequence. Recall that the module-valued functors $\underline{M}, I^n, \underline{P}$ each have property $(*)$, see Lemma \ref{lemma-tangent-injective} and the remarks preceding it. By splitting $0 \to \underline{M} \to I^\bullet$ into short exact sequences we find that each of the functors $\Im(d^{n - 1}) = \Ker(d^n) \subset I^n$ has property $(*)$ by Lemma \ref{lemma-exactness-permanence} and also that $I^{i - 1} \times_{\Ker(d^i)} \underline{P}$ has property $(*)$. \medskip\noindent Thus we may assume the Yoneda extension is given as $$ E : 0 \to \underline{M} \to F_{i - 1} \to \ldots \to F_0 \to \underline{P} \to 0 $$ where each of the module-valued functors $F_j$ has property $(*)$. Set $G_j(B) = TF_j(B, B)$ viewed as a $B$-module via the {\it second} $B$-module structure defined in Lemma \ref{lemma-exactness-implies}. Since $TF_j$ is a functor on pairs we see that $G_j$ is a module-valued functor on $\textit{Alg}_A$. Moreover, since $E$ is an exact sequence the sequence $G_{j + 1} \to G_j \to G_{j - 1}$ is exact (see remark preceding Lemma \ref{lemma-exactness-permanence}). Observe that $T\underline{M}(B, B) = M \otimes_A B = \underline{M}(B)$ and that the two $B$-module structures agree on this. Thus we obtain a Yoneda extension $$ E' :  0 \to \underline{M} \to G_{i - 1} \to \ldots \to G_0 \to \underline{P} \to 0 $$ Moreover, the canonical maps $$ F_j(B) = B \otimes_B F_j(B) \longrightarrow TF_j(B, B) = G_j(B) $$ of Lemma \ref{lemma-tangent-functor} (\ref{item-mult-map}) are $B$-linear by Lemma \ref{lemma-exactness-implies} (\ref{item-mult-map-linear}) and functorial in $B$. Hence a map $$ \xymatrix{ 0 \ar[r] & \underline{M} \ar[r] \ar[d]^1 & F_{i - 1} \ar[r] \ar[d] & \ldots \ar[r] & F_0 \ar[r] \ar[d] & \underline{P} \ar[r] \ar[d]^1 & 0 \\ 0 \ar[r] & \underline{M} \ar[r] & G_{i - 1} \ar[r] & \ldots \ar[r] & G_0 \ar[r] & \underline{P} \ar[r] & 0 } $$ of Yoneda extensions. In particular we see that $E$ and $E'$ have the same class in $\Ext^i_\mathcal{P}(\underline{P}, \underline{M})$ by the lemma on Yoneda Exts mentioned above. Finally, let $N$ be a $A$-module of finite presentation. Then we see that $$ 0 \to T\underline{M}(A, N) \to TF_{i - 1}(A, N) \to \ldots \to TF_0(A, N) \to T\underline{P}(A, N) \to 0 $$ is exact. By Lemma \ref{lemma-exactness-implies} (\ref{item-tangent-right-exact}) with $B = A$ this translates into the exactness of the sequence of $A$-modules $$ 0 \to M \otimes_A N \to G_{i - 1}(A) \otimes_A N \to \ldots \to G_0(A) \otimes_A N \to P \otimes_A N \to 0 $$ Hence the sequence of $A$-modules $0 \to M \to G_{i - 1}(A) \to \ldots \to G_0(A) \to P \to 0$ is universally exact, in the sense that it remains exact on tensoring with any finitely presented $A$-module $N$. Let $K = \Ker(G_0(A) \to P)$ so that we have exact sequences $$ 0 \to K \to G_0(A) \to P \to 0 \quad\text{and}\quad G_2(A) \to G_1(A) \to K \to 0 $$ Tensoring the second sequence with $N$ we obtain that $K \otimes_A N = \Coker(G_2(A) \otimes_A N \to G_1(A) \otimes_A N)$. Exactness of $G_2(A) \otimes_A N \to G_1(A) \otimes_A N \to G_0(A) \otimes_A N$ then implies that $K \otimes_A N \to G_0(A) \otimes_A N$ is injective. By Algebra, Theorem \ref{algebra-theorem-universally-exact-criteria} this means that the $A$-module extension $0 \to K \to G_0(A) \to P \to 0$ is exact, and because $P$ is assumed of finite presentation this means the sequence is split, see Algebra, Lemma \ref{algebra-lemma-universally-exact-split}. Any splitting $P \to G_0(A)$ defines a map $\underline{P} \to G_0$ which splits the surjection $G_0 \to \underline{P}$. Thus the Yoneda extension $E'$ is equivalent to the trivial Yoneda extension and we win.