Solution
Since the pulley is ideal and massless, the two string tensions are equal. Therefore, for the pulley,
$$ F=2T $$ or $$ \boxed{T=\frac{F}{2}}. $$The tension tends to pull both blocks towards the right relative to the platform. Hence friction on each block, whenever required, acts towards the left. The corresponding reaction friction on the platform acts towards the right.
Let $f_1$ be the friction between A and P and $f_2$ the friction between B and P. We shall use only their magnitudes.
1. First suppose both blocks remain at rest relative to the platform
Then A, B and P all have acceleration $$ a=2\,\text{m/s}^2. $$
The total mass of the complete system is $$ 4+1+1=6\,\text{kg}. $$ The only external horizontal force on the complete system is $F$. Therefore, $$ F=(6)(2) $$ $$ \boxed{F=12\,\text{N}}. $$ Hence the tension would be $$ T=\frac{F}{2}=6\,\text{N}. $$
Now consider block B.
Tension acts towards the right and friction acts towards the left: $$ T-f_2=m_Ba. $$ Therefore, $$ 6-f_2=(1)(2), $$ giving $$ \boxed{f_2=4\,\text{N}}. $$
But the maximum possible static friction on B is $$ f_{2,\max} = \mu_s m_Bg = 0.16(1)(10) = 1.6\,\text{N}. $$
2. Since B slips, friction between B and P is kinetic
Therefore, $$ f_2=\mu_km_Bg $$ $$ f_2=(0.10)(1)(10) $$ $$ \boxed{f_2=1\,\text{N}}. $$
This friction acts towards the left on B and hence towards the right on P.
Now consider the platform P.
The only horizontal forces on it are the friction forces due to A and B. Since its mass is $1\,\text{kg}$ and its acceleration is $2\,\text{m/s}^2$, $$ f_1+f_2=m_Pa_P. $$ Thus, $$ f_1+1=(1)(2), $$ so $$ \boxed{f_1=1\,\text{N}}. $$
We must now check whether this amount of friction can be provided statically between A and P.
For block A, $$ f_{1,\max} = \mu_s m_Ag = 0.16(4)(10) = 6.4\,\text{N}. $$ Since $$ 1\,\text{N}<6.4\,\text{N}, $$ the required friction is easily possible.
3. Find the force $F$
For block A, tension acts towards the right and friction $f_1=1\,\text{N}$ acts towards the left.
$$ T-f_1=m_Aa_A $$ $$ T-1=(4)(2) $$ $$ T=9\,\text{N}. $$Since $F=2T$,
$$ \boxed{F=18\,\text{N}}. $$4. Acceleration of block B
For B, $$ T-f_2=m_Ba_B. $$ Therefore, $$ 9-1=(1)a_B, $$ which gives $$ \boxed{a_B=8\,\text{m/s}^2}. $$
Thus B accelerates faster than the platform and therefore slides towards the right relative to it, exactly as assumed.
Final Answer
$$ \boxed{F=18\,\text{N}} $$For block A:
$$ \boxed{f_A=1\,\text{N}\ \text{towards left}} $$ $$ \boxed{a_A=2\,\text{m/s}^2\ \text{towards right}} $$A remains at rest relative to the platform.
For block B:
$$ \boxed{f_B=1\,\text{N}\ \text{towards left}} $$ $$ \boxed{a_B=8\,\text{m/s}^2\ \text{towards right}} $$B slips towards the right relative to the platform.