For $x\in X$, $y\in Y$, $\lambda_1\in X^*$, $\lambda_2\in Y^*$, we have the inequality
\begin{align}
|\lambda_1(x)+\lambda_2(y)|&\leq|\lambda_1(x)|+|\lambda_2(y)|\leq \|\lambda_1\|\,\|x\|+\|\lambda_2\|\,\|y\|\\ \ \\ &\leq (\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/q}\,(\|x\|^p+\|y\|^p)^{1/p}
\end{align}
(where we used Hölder in the last $\leq$). This shows that
$$
\|T(\lambda_1,\lambda_2)\| \leq \|\lambda_1,\lambda_2\|_q.
$$
Now fix $\varepsilon>0$ and let $x'\in X$, $y'\in Y$ with $\|x'\|=\|y'\|=1$, and $\|\lambda_1\|\leq|\lambda_1(x')|-\varepsilon$, $\|\lambda_2\|\leq|\lambda_2(y)|-\varepsilon$. Let
$$
x=\frac{\|\lambda_1\|^{q-1}}{\|(\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\,x', \ \ \
y=\frac{\|\lambda_2\|^{q-1}}{\|(\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\,y'.
$$
Then
$$
\|(x,y)\|_p^p=\frac{\|\lambda_1\|^{p(q-1)}}{\|\lambda_1\|^q+\|\lambda_2\|^q}+\frac{\|\lambda_2\|^{p(q-1)}}{\|\lambda_1\|^q+\|\lambda_2\|^q}=\frac{\|\lambda_1\|^q+\|\lambda_2\|^q}{\|\lambda_1\|^q+\|\lambda_2\|^q}=1,
$$
so $\|(x,y)\|_p=1$. And
\begin{align}
T(\lambda_1,\lambda_2)(x,y)&=\lambda_1(x)+\lambda_2(y)=\frac{\|\lambda_1\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\,|\lambda_1(x')|+\frac{\|\lambda_2\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\,|\lambda_2(y')|\\ \ \\
&\geq\frac{\|\lambda_1\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\,(\|\lambda_1\|-\varepsilon)+\frac{\|\lambda_2\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\,(\|\lambda_2\|-\varepsilon)\\ \ \\
&=\frac{\|\lambda_1\|^q+\|\lambda_2\|^q}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}-\varepsilon\,\frac{\|\lambda_1\|^{q-1}+\|\lambda_2\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\\ \ \\
&=(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/q}-\varepsilon\,\frac{\|\lambda_1\|^{q-1}+\|\lambda_2\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\\ \ \\
&=\|(\lambda_1,\lambda_2)\|_q-\varepsilon\,\frac{\|\lambda_1\|^{q-1}+\|\lambda_2\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\\ \ \\.
\end{align}
As $\varepsilon$ was arbitrary, we get that $\|T(\lambda_1,\lambda_2)\|\geq\|(\lambda_1,\lambda_2)\|_q$.
So $\|T(\lambda_1,\lambda_2)\|=\|(\lambda_1,\lambda_2)\|_q$ for all $\lambda_1,\lambda_2$, i.e. $T$ is an isometry.
It remains to show that $T$ is onto, but I assume you can do that.