Answer
$\text{ Crash into the Earth.}$
Work Step by Step
As the author told us, the two satellites will move as one unit after the collision. We need to find the final velocity after the collision, so we can use the conservation of momentum.
But first, we need to find this orbit's speed.
$$\sum F=F_G=\dfrac{GM_E\color{red}{\bf\not}m}{R_o^2}=\color{red}{\bf\not}ma_r$$
$$\dfrac{GM_E }{R_o^{\color{red}{\bf\not}2}}=\dfrac{v_i^2}{\color{red}{\bf\not}R_o}$$
where $v_i$ is the initial speed of the two satellites before the collision, and $R_o=R_E+h$ where $R_E$ is the earth's radius and $h$ is the height of the orbit from the earth's surface.
Thus,
$$v_i=\sqrt{\dfrac{GM_E }{R_E+h}}\tag 1$$
The momentum is conserved before and after the collision, so
$$p_i=p_f$$
$$m_1v_{i}+m_2(-v_i)=(m_1+m_2)v_f$$
where $(-v_i)$ is due to direction since they are moving in opposite directions, and $m_1=400$ kg, $m_2=100$ kg.
$$v_i\left[m_1 -m_2 \right]=(m_1+m_2)v_f$$
Hence,
$$v_f=\dfrac{v_i\left[m_1 -m_2 \right]}{m_1+m_2}$$
Plugging from (1);
$$v_f=\dfrac{\sqrt{\dfrac{GM_E }{R_E+h}}\left[m_1 -m_2 \right]}{m_1+m_2}$$
Plugging the known;
$$v_f=\dfrac{\sqrt{\dfrac{(6.67\times 10^{-11})(5.98\times 10^{24})}{(6.37\times 10^6)+(1000\times 10^3)}}\left[400-100\right]}{400+100}$$
$$v_f=\bf4414\;\rm m/s$$
After the collision, the two satellites as one unit will move on an elliptical orbit around the earth that starts at a speed of 4414 m/s at a distance of 1000 km (which was their initial orbit) above the earth's surface. So we need to find the distance between them and the Earth at the closest point to earth. See the second figure below.
We need to find $h_1$ and to find $h_1$ we need to find $v_1$ which is their speed at the closest point to the Earth.
We can use the conservation of energy
$$K_i+U_{ig}=K_f+U_{fg}$$
$$\frac{1}{2}\color{red}{\bf\not}(m_1+m_2)v_f^2+\dfrac{-GM_E\color{red}{\bf\not}(m_1+m_2)}{R_E+h}=\frac{1}{2}\color{red}{\bf\not}(m_1+m_2)v_1^2+\dfrac{-GM_E\color{red}{\bf\not}(m_1+m_2)}{R_E+h_1}$$
If $h_1$ is negative then the two satellites will hit the ground.
$$\frac{1}{2} v_f^2+\dfrac{-GM_E }{R_E+h}=\frac{1}{2} v_1^2+\dfrac{-GM_E }{R_E+h_1}$$
Hence,
$$ v_f^2-\dfrac{ 2GM_E }{R_E+h}+ \dfrac{ 2GM_E }{R_E+h_1}= v_1^2\tag 2$$
Now we know that the linear momentum is conserved as well, so
$$L_i=L_1$$
$$r_i\color{red}{\bf\not}(m_1+m_2)v_f=r_i\color{red}{\bf\not}(m_1+m_2)v_1$$
$$(R_E+h) v_f=(R_E+h_1)v_1$$
Hence,
$$v_1=\dfrac{(R_E+h) v_f}{(R_E+h_1) }$$
Plugging into (2);
$$ v_f^2-\dfrac{ 2GM_E }{R_E+h}+ \dfrac{ 2GM_E }{R_E+h_1}= \left(\dfrac{(R_E+h) v_f}{(R_E+h_1) }\right)^2 $$
$$ v_f^2-\dfrac{ 2GM_E }{R_E+h}+ \dfrac{ 2GM_E }{R_E+h_1}= \dfrac{(R_E+h)^2 v_f^2}{(R_E+h_1)^2 } $$
$$ v_f^2-\dfrac{ 2GM_E }{R_E+h}= \dfrac{(R_E+h)^2 v_f^2}{(R_E+h_1)^2 } - \dfrac{ 2GM_E }{R_E+h_1}$$
$$ \dfrac{ (R_E+h)v_f^2-2GM_E }{R_E+h}= \dfrac{(R_E+h)^2 v_f^2-2GM_E(R_E+h_1)}{(R_E+h_1)^2 } $$
Let $R_E+h_1=x$ and solve for it.
$$ \dfrac{ (R_E+h)v_f^2-2GM_E }{R_E+h}= \dfrac{(R_E+h)^2 v_f^2-2GM_Ex}{x^2 } $$
$$ \left[\dfrac{ (R_E+h)v_f^2-2GM_E }{R_E+h}\right]x^2= (R_E+h)^2 v_f^2-2GM_Ex $$
$$ \left[\dfrac{ (R_E+h)v_f^2-2GM_E }{R_E+h}\right]x^2+\left[2GM_E\right]x - (R_E+h)^2 v_f^2=0 $$
Plugging the known;
$$ \left[\dfrac{ ([6.37\times 10^6]+[1000\times 10^3])(4414)^2-2(6.67\times 10^{-11})(5.98\times 10^{24}) }{[6.37\times 10^6]+[1000\times 10^3]}\right]x^2+\left[2(6.67\times 10^{-11})(5.98\times 10^{24})\right]x - ([6.37\times 10^6]+[1000\times 10^3])^2(4414)^2=0 $$
$$(-8.875\times10^7)x^2+(7.97732\times 10^14)x-(1.05828\times 10^21)=0$$
Hence,
$$x=1.61779\times10^6\;{\rm m\;\;\;\;\;Or, }\;\;\;\;\;\;x=7.37074\times10^6\;\rm m$$
We can see that the second root is the initial position of the two satellites just after the collision. So it is not the one we seeking for,
Let's work with the first root.
$$x=R_E+h_1=1.61779\times10^6\;\rm m$$
Hence,
$$h_1=1.61779\times10^6\;\rm m-R_E=1.61779\times10^6\;\rm m-(6.37\times 10^6)\;\rm m$$
$$h_1=\bf -4.75\times 10^6\;\rm m$$
This means that they will hit the Earth and crash into the Earth.
