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I claim $K^+(V^\lambda)$, for any $\lambda$ (I switched the notation so not to be confused with the von Neumann universe) is equiconsistent with the schema "$ORD$ is Mahlo" (Not to be confused with full stationarity, which could be called "$Ord$ is Mahlo" if you really wanted to distinguish them). First off each $V^\lambda$ is a Gothrendick universe, and so of the form $V_\kappa$ for inaccessible $\kappa$. Second of all, $V_\kappa\prec W$, where $W=\{x|x=x\}$. To see this, suppose $\exists x(\phi(x,x_0...x_n))$ where $\phi(x,x_0...x_n)$ is absolute.

Then $\exists x(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi^{V^\lambda}(x,x_0...x_n))$ if and only if $V^\lambda\vDash\exists x(\phi(x,x_0...x_n))$. Note that at no point do we use reflection for a formula that uses $V^\lambda$. Therefore $K^+(V^\lambda)$ proves that there exists a reflection cardinal, and so the consistency strength of $K^+(V^\lambda)\ge$ the consistency strength of "$ORD$ is Mahlo."

Then, suppose $ORD$ is Mahlo. Then there exists a proper class of reflecting cardinals, and if we take $V^\lambda=V_\kappa$ for reflecting $\kappa$, we get $K^+(V^\lambda)$. Therefore the consistency strength of "$ORD$ is Mahlo"$\ge$ the consistency strength of $K^+(V^\lambda)$, and so the consistency of "$ORD$ is Mahlo"$=$ the consistency strength of $K^+(V^\lambda)$.

I claim $K^+(V^\lambda)$, for any $\lambda$ (I switched the notation so not to be confused with the von Neumann universe) is equiconsistent with the schema "$ORD$ is Mahlo" (Not to be confused with full stationarity, which could be called "$Ord$ is Mahlo" if you really wanted to distinguish them). First off each $V^\lambda$ is a Grothendieck universe, and so of the form $V_\kappa$ for inaccessible $\kappa$. Second of all, $V_\kappa\prec W$, where $W=\{x|x=x\}$. To see this, suppose $\exists x(\phi(x,x_0...x_n))$ where $\phi(x,x_0...x_n)$ is absolute.

Then $\exists x(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi^{V^\lambda}(x,x_0...x_n))$ if and only if $V^\lambda\vDash\exists x(\phi(x,x_0...x_n))$. Note that at no point do we use reflection for a formula that uses $V^\lambda$. Therefore $K^+(V^\lambda)$ proves that there exists a reflection cardinal, and so the consistency strength of $K^+(V^\lambda)\ge$ the consistency strength of "$ORD$ is Mahlo."

Then, suppose $ORD$ is Mahlo. Then there exists a proper class of reflecting cardinals, and if we take $V^\lambda=V_\kappa$ for reflecting $\kappa$, we get $K^+(V^\lambda)$. Therefore the consistency strength of "$ORD$ is Mahlo"$\ge$ the consistency strength of $K^+(V^\lambda)$, and so the consistency of "$ORD$ is Mahlo"$=$ the consistency strength of $K^+(V^\lambda)$.

I claim $K^+(V^\lambda)$, for any $\lambda$ (I switched the notation so not to be confused with the von Neumann universe) is equiconsistent with the schema "$ORD$ is Mahlo" (Not to be confused with full stationarity, which could be called "$Ord$ is Mahlo" if you really wanted to distinguish them). First off each $V^\lambda$ is a Gothrendick universe, and so of the form $V_\kappa$ for inaccessible $\kappa$. Second of all, $V_\kappa\prec W$, where $W=\{x|x=x\}$. To see this, suppose $\exists x(\phi(x,x_0...x_n))$ where $\phi(x,x_0...x_n)$ is absolute.

Then $\exists x(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi^{V^\lambda}(x,x_0...x_n))$ if and only if $V^\lambda\vDash\exists x(\phi(x,x_0...x_n))$. Note that out no point do we use reflection for a formula that uses $V^\lambda$. Therefore $K^+(V^\lambda)$ proves that there exists a reflection cardinal, and so the consistency strength of $K^+(V^\lambda)\ge$ the consistency strength of "$ORD$ is Mahlo."

Then, suppose $ORD$ is Mahlo. Then there exists a proper class of reflecting cardinals, and if we take $V^\lambda=V_\kappa$ for reflecting $\kappa$, we get $K^+(V^\lambda)$. Therefore the consistency strength of "$ORD$ is Mahlo"$\ge$ the consistency strength of $K^+(V^\lambda)$, and so the consistency of "$ORD$ is Mahlo"$=$ the consistency strength of $K^+(V^\lambda)$.

I claim $K^+(V^\lambda)$, for any $\lambda$ (I switched the notation so not to be confused with the von Neumann universe) is equiconsistent with the schema "$ORD$ is Mahlo" (Not to be confused with full stationarity, which could be called "$Ord$ is Mahlo" if you really wanted to distinguish them). First off each $V^\lambda$ is a Gothrendick universe, and so of the form $V_\kappa$ for inaccessible $\kappa$. Second of all, $V_\kappa\prec W$, where $W=\{x|x=x\}$. To see this, suppose $\exists x(\phi(x,x_0...x_n))$ where $\phi(x,x_0...x_n)$ is absolute.

Then $\exists x(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi^{V^\lambda}(x,x_0...x_n))$ if and only if $V^\lambda\vDash\exists x(\phi(x,x_0...x_n))$. Note that at no point do we use reflection for a formula that uses $V^\lambda$. Therefore $K^+(V^\lambda)$ proves that there exists a reflection cardinal, and so the consistency strength of $K^+(V^\lambda)\ge$ the consistency strength of "$ORD$ is Mahlo."

Then, suppose $ORD$ is Mahlo. Then there exists a proper class of reflecting cardinals, and if we take $V^\lambda=V_\kappa$ for reflecting $\kappa$, we get $K^+(V^\lambda)$. Therefore the consistency strength of "$ORD$ is Mahlo"$\ge$ the consistency strength of $K^+(V^\lambda)$, and so the consistency of "$ORD$ is Mahlo"$=$ the consistency strength of $K^+(V^\lambda)$.

I claim $K^+(V^\lambda)$, for any $\lambda$ (I switched the notation so not to be confused with the von Neumann universe) is equiconsistent with the schema "$ORD$ is Mahlo" (Not to be confused with full stationarity, which could be called "$Ord$ is Mahlo" if you really wanted to distinguish them). First off each $V^\lambda$ is a Gothrendick universe, and so of the form $V_\kappa$ for inaccessible $\kappa$. Second of all, $V_\kappa\prec W$, where $W\{x|x=x\}$. To see this, suppose $\exists x(\phi(x,x_0...x_n))$ where $\phi(x,x_0...x_n)$ is absolute.

Then $\exists x(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi^{V^\lambda}(x,x_0...x_n))$ if and only if $V^\lambda\vDash\exists x(\phi(x,x_0...x_n))$. Note that out no point do we use reflection for a formula that uses $V^\lambda$. Therefore $K^+(V^\lambda)$ proves that there exists a reflection cardinal, and so the consistency strength of $K^+(V^\lambda)\ge$ the consistency strength of "$ORD$ is Mahlo."

Then, suppose $ORD$ is Mahlo. Then there exists a proper class of reflecting cardinals, and if we take $V^\lambda=V_\kappa$ for reflecting $\kappa$, we get $K^+(V^\lambda)$. Therefore the consistency strength of "$ORD$ is Mahlo"$\ge$ the consistency strength of $K^+(V^\lambda)$, and so the consistency of "$ORD$ is Mahlo"$=$ the consistency strength of $K^+(V^\lambda)$.

I claim $K^+(V^\lambda)$, for any $\lambda$ (I switched the notation so not to be confused with the von Neumann universe) is equiconsistent with the schema "$ORD$ is Mahlo" (Not to be confused with full stationarity, which could be called "$Ord$ is Mahlo" if you really wanted to distinguish them). First off each $V^\lambda$ is a Gothrendick universe, and so of the form $V_\kappa$ for inaccessible $\kappa$. Second of all, $V_\kappa\prec W$, where $W=\{x|x=x\}$. To see this, suppose $\exists x(\phi(x,x_0...x_n))$ where $\phi(x,x_0...x_n)$ is absolute.

Then $\exists x(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi^{V^\lambda}(x,x_0...x_n))$ if and only if $V^\lambda\vDash\exists x(\phi(x,x_0...x_n))$. Note that out no point do we use reflection for a formula that uses $V^\lambda$. Therefore $K^+(V^\lambda)$ proves that there exists a reflection cardinal, and so the consistency strength of $K^+(V^\lambda)\ge$ the consistency strength of "$ORD$ is Mahlo."

Then, suppose $ORD$ is Mahlo. Then there exists a proper class of reflecting cardinals, and if we take $V^\lambda=V_\kappa$ for reflecting $\kappa$, we get $K^+(V^\lambda)$. Therefore the consistency strength of "$ORD$ is Mahlo"$\ge$ the consistency strength of $K^+(V^\lambda)$, and so the consistency of "$ORD$ is Mahlo"$=$ the consistency strength of $K^+(V^\lambda)$.