First we put it in the notation of Mathematica $K(k)$ is $K(k^2)$.
So our function will be
$$f(k)= k K(k^2)\sinh\Bigl(\frac{\pi}{2}\frac{K(1-k^2)}{K(k^2)}\Bigr).$$
Now we change variables (W486, Whittaker, Watson p.~486) .
$$k=\frac{\vartheta_2^2(q)}{\vartheta_3^2(q)} \quad (*)$$
Where$\newcommand\Z{\mathbb{Z}}$
$$\vartheta_2(q)=2q^{\frac14}(1+q^2+q^6+\cdots)= \sum_{n\in\Z}q^{(n-\frac12)^2}=
2q^{\frac14}\prod_{n=1}^\infty\{(1-q^{2n})(1+q^{2n})^2\}$$
$$\vartheta_3(q)=1+2q+2q^4+2q^9+\cdots=\sum_{n\in\Z}q^{n^2}=
\prod_{n=1}^\infty\{(1-q^{2n})(1+q^{2n-1})^2\}$$
$$\vartheta_4(q)=1-2q+2q^4-2q^9+\cdots=\sum_{n\in\Z}(-1)^nq^{n^2}=
\prod_{n=1}^\infty\{(1-q^{2n})(1-q^{2n-1})^2\}$$
The function of $q$ in (*) is differentiable and increasing on $(0,1)$
it is $0$ in $0$ and $1$ in $1$.
(we shall write $\vartheta_j$ to denote $\vartheta_j(q)$).
Since (W467)
$$\vartheta_2^4+\vartheta_4^4=\vartheta_3^4$$
we have
$$1-k^2=1-\frac{\vartheta_2^4}{\vartheta_3^4}=\frac{\vartheta_4^4}{\vartheta_3^4}$$
The interesting thing about this change of variables is that
$$K(k^2)=K\Bigl(\frac{\vartheta_2^4}{\vartheta_3^4}\Bigr)=\frac{\pi}{2}\vartheta_3^2,\qquad
K(1-k^2)=K\Bigl(\frac{\vartheta_4^4}{\vartheta_3^4}\Bigr)=\frac{\log(1/q)}{2}\vartheta_3^2.$$

Now our function is
$$k K(k^2)\sinh\Bigl(\frac{\pi}{2}\frac{K(1-k^2)}{K(k^2)}\Bigr)
=\frac{\pi}{4}\Bigl(\frac{1}{\sqrt{q}}-\sqrt{q}\Bigr) \vartheta_2^2$$
that must be decreasing in $q$.

We add to the above some comments:

We need to show that $f(q):=(1-q)\vartheta_2^2/4\sqrt{q}$ is decreasing for $0 < q < 1 $. But

$$f(q)=(1-q)\prod_{n=1}^\infty (1-q^{2n})^2(1+q^{2n})^4=
(1-q)\prod_{n=1}^\infty (1-q^{4n})^2(1+q^{2n})^2$$
This is the same to prove that the logarithmic derivative is negative
$$-\frac{1}{1-q}-\sum_{n=1}^\infty\frac{8nq^{4n-1}}{1-q^{4n}}+\sum_{n=1}^\infty
\frac{4nq^{2n-1}}{1+q^{2n}}$$
multiply this by $q>0$ and expand in series
$$-\sum_{m=1}^\infty q^m-\sum_{n=1}^\infty \sum_{k=1}^\infty 8nq^{4nk}+
\sum_{n=1}^\infty\sum_{k=1}^\infty (-1)^{k+1}4nq^{2nk}$$

To show that this is negative observe that the only positive terms are those in
the third sum with $k=2j+1$ odd, we will pair this term with that in the first
sum corresponding to the same $n$ and $k=j$, these two terms are
$$-8nq^{4nj}+4nq^{2n(2j+1)}=-4nq^{4nj}(2-q^{2n})<0.$$
This leaves only the terms with $j=0$ without pair.
The remaining positive terms adds to
$$\sum_{n=1}^\infty 4nq^{2n}=\frac{4q^2}{(1-q^2)^2}.$$
These terms we compensate with the first sum
$$-\frac{q}{1-q}+\frac{4q^2}{(1-q^2)^2}=-\frac{q(1+q)(1-q^2)-4q^2}{(1-q^2)^2}$$
This is negative for $ 0 < q < 0.295598 $.

There is an intrinsic difficulty to treat larger values of $q$.

I propose to use the modularity of the theta function:

We have the equality considering $\vartheta_j$ as functions of $q$
$$\vartheta_2(e^{-\frac{\pi}{x}})=\sqrt{x}\vartheta_4(e^{-\pi x})$$
It follows that putting $q=e^{-\frac{\pi}{x}}$
$$\frac{1}{4}\Bigl(\frac{1}{\sqrt{q}}-\sqrt{q}\Bigr)\vartheta_2^2(e^{-\pi/x})=
\frac{x}{2}\sinh\frac{\pi}{2x}\, \vartheta_4^2(e^{-\pi x}).$$
We must show this function is decreasing
Since
$$\vartheta_4(e^{-\pi x}) = \prod_{n=1}^\infty
(1-e^{-2\pi n x})(1-e^{-\pi(2n-1)x})^2$$
is almost $1$ for $x$ near infinite, we only have to show that for $x$ large $\frac{x}{2}\sinh\frac{\pi}{2x}$ is decreasing

In this way we are proving that our function decrease when the initial $q$ is near 1.
This was the difficult part before.

This strategy must be sufficient.