$(X \oplus_p Y)^*$ isometric to $(X^*\oplus_q Y^*)$

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 $\|\lambda_1\|\leq|\lambda_1(x')|-\varepsilon$, $\|\lambda_2\|\leq|\lambda_2(y')|-\varepsilon$. By multiplying each by an appropriate complex number with absolute value 1, we may assume that $\lambda_1(x')\geq0$, $\lambda_2(y')\geq0$. 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.