The problem with this question is that static friction and kinetic friction are not fundamental forces in any way-- they're purely phenomenological names used to explain observed behavior. "Static friction" is a term we use to describe the observed fact that it usually takes more force to set an object into motion than it takes to keep it moving once you've got it started.
So, with that in mind, ask yourself how you could measure the relative sizes of static and kinetic friction. If the coefficient of static friction is greater than the coefficient of kinetic friction, this is an easy thing to do: once you overcome the static friction, the frictional force drops. So, you pull on an object with a force sensor, and measure the maximum force required before it gets moving, then once it's in motion, the frictional force decreases, and you measure how much force you need to apply to maintain a constant velocity.
What would it mean to have kinetic friction be greater than static friction? Well, it would mean that the force required to keep an object in motion would be greater than the force required to start it in motion. Which would require the force to go up at the instant the object started moving. But that doesn't make any sense, experimentally-- what you would see in that case is just that the force would increase up to the level required to keep the object in motion, as if the coefficients of static and kinetic friction were exactly equal.
So, common sense tells us that the coefficient of static friction can never be less than the coefficient of kinetic friction. Having greater kinetic than static friction just doesn't make any sense in terms of the phenomena being described.
(As an aside, the static/kinetic coefficient model is actually pretty lousy. It works as a way to set up problems forcing students to deal with the vector nature of forces, and allows some simple qualitative explanations of observed phenomena, but if you have ever tried to devise a lab doing quantitative measurements of friction, it's a mess.)
In my opinion, the requirement that the string be nonextensible creates conceptual issues.
On the one hand, it is stated that the string is nonextensible. On the other hand, it is stated on the diagram the "External force $F$ is applied such that the block remains at rest". The problem is if the string is nonextensible, and initially has no slack, then the block cannot move to the left, i.e., it will always be "at rest", regardless of the force $F$ to the left.
But more importantly, it makes it difficult to explain (1) why the static friction force counters the applied force before the tension in the string and (2) why when equilibrium is reached the friction force no longer exists.
To facilitate the answers to these questions I will replace the string with an ideal spring (see the figures below). An ideal spring, like the string, is massless. But unlike the string, it is extensible to the degree allowed by the spring constant. The magnitude of the tension in the spring is equals the magnitude of its restoring force, or $T=k\Delta x$ where $k$ is the spring constant and $\Delta x$ is its extension beyond the "relaxed" stated.
Now consider the following where the block is considered a rigid (nonextensible) body:
The spring, like the string, is initially relaxed so there is no tension. FIG 1 shows the block with no external force and the relaxed spring attached to the wall
In Fig 2 we gradually apply an increasing external force $F$ that is less than the maximum static friction force. Since the block cannot move, the spring cannot extend and thus the tension in the spring is still zero. This explains, for a physically real scenario, why the applied force $F$ is countered first by the static friction force.
In FIG 3 the applied force reaches the maximum static friction force and the friction force becomes kinetic friction, which is generally considered constant. Since kinetic friction is usually less than static friction, if the applied force $F$ is maintained at the value of the maximum static friction there will initially be a net force to the left causing the block to move to the left. (Note that actual value of the friction force during the transition from static to kinetic is undefined for the standard model of friction). At the same time, however, the spring extends creating an opposing tension force. So during this phase before the extension of the spring is a maximum, we have
$$F-f_{k}> T$$
$$\mu_{s}mg-\mu_{k}mg>k\Delta x$$
and the block is moving to the left.
- When the extension of the spring is such that the tension in the spring equals the applied force $F$, it's extension is maximized and we have
$$F=T=k\Delta x_{max}$$
Substituting into the first equation,
$$f_{k}=0$$
Meaning there is no net force for friction to oppose.
Note that in this example, the stiffer the spring (the greater $k$ is) the less the block needs to move before the tension equals the applied force, i.e., the quicker the tension rises. The nonextensible string is simply a spring with an infinite $k$.
Hope this helps.
Best Answer
If the inital force $Kx$ from the spring on block A is less than or equal to the maximum static friction between the two blocks then friction from block B will exert a force $F$ to the right on block A that is sufficient to prevent relative motion between the blocks. Note that no relative motion does not means that block A is at equilibrium - if block B accelerates then block A must accelerate too.
Since Block B exerts a force $F$ to the right on Block A, then Block A will exert an equal and opposite force $F$ to the left on block B. So block B will accelerate at a rate $a= \frac F {m_B}$ to the left. Since block A cannot move relative to block B due to the friction between them, block A must also accelerate at a rate $a$ to the left as well. So resolving the forces on block A we have
$\displaystyle Kx - F = m_Aa = F \frac {m_A}{m_B} \\ \displaystyle \Rightarrow Kx = F \frac {m_A+m_B}{m_B} \\ \displaystyle \Rightarrow F = \frac {Kxm_B}{m_A+m_B} \\ \displaystyle \Rightarrow a = \frac {Kx}{m_A+m_B}$
In other words, the two blocks will move as if they were a single block with mass $m_A+m_B$.
I'll let you work out what happens as the blocks start to move to the left and $x$ decreases.