Let $1<p<\infty$ and $\phi\in L^{\infty}(\Omega)$ be a given positive and bounded function, where $\Omega\subset\mathbb{R}^N$, $N\geq 2$ is bounded. Let $\mu>0$ and $\lambda>0$ be given constants and $\gamma$ be a positive continuous function over $\Omega$ such that $0<\gamma(x)<1$ for all $x\in\Omega$. Suppose $h:[0,\infty)\to(0,\infty)$ be a continuous and non-decreasing function such that $$ \lim_{t\to 0}t^{-\gamma(x)}h(t)=\infty. $$ Then there exists a constant $a_{\lambda}>0$ (independent of $\mu,x$) such that $$ \mu(a_{\lambda}\phi)^{p-1}\leq\lambda (a_{\lambda}\phi)^{-\gamma(x)}h(a_{\lambda}\phi)\text{ in }\Omega. $$
I can see that, if we fix $x=x_0\in\Omega$, then we can get $a_{\lambda}(x_0)$ depedning on $x_0$ such that $$ \mu(a_{\lambda}\phi)^{p-1}\leq\lambda (a_{\lambda}\phi)^{-\gamma(x)}h(a_{\lambda}\phi) $$ holds at the point $x=x_0$. But unable to understand, how to obtain $a_{\lambda}$ to be a constant independent of $x,\mu$. Can someone kindly help. Thanks.
Let $$ f(x,t) = t^{p-1+\gamma(x)}(h(t))^{-1}.$$ For $t<1$, $$ \vert f(x,t) \vert \leqslant t^{p-1} (h(t))^{-1} \to 0 $$ as $t\to 0^+$ since $h$ is continuous, $h(0)>0$, and $p>1$. Hence, $f(x,t) \to 0$ as $t\to 0^+$ uniformly in $x$. This means that there exists some $\delta >0$ independent of $x$ such that if $t<\delta$ then$$ f(x,t)<\lambda / \mu . $$ Let $a = \frac\delta{2 \| \phi \|_{L^\infty}}$. Then $$ a \phi = \frac{\delta \phi}{2 \| \phi \|_{L^\infty}(\Omega)} \leqslant \frac\delta2<\delta$$ so $$f(x,a\phi) <\lambda / \mu $$ which is the inequality you are looking for.
As it stands this $a$ does not depend on $x$ but it does depend on $\lambda$ and $\mu$ (in fact, it depends only on $h$, $p$, and the ratio $\lambda/\mu$). It is impossible to find such an $a$ that does not depend on $\mu$. Indeed, if there were then you could send $\mu \to \infty$ in your inequality and get a contradiction.