Pumping Lemma states, if $n$ is the length of any string of length $s,$ the sequence of state $q_{1},q_{3},q_{20},q_{9},........q_{13}$ has length $n+1.$ Because we know, $n$ is atleast $p$ (where $p$ is pumping length), and we know $n+1$ is grater than $p,$ the number of states in the language $M.$ The sequence must contain a repeated state, i.e. fancy name of Pegion-hole Principle
If the string M divide in $xyz$, then $x$ goes from $r_{1}$ to $r_{j}$, $y$ takes $M$ from $r_{j}$ to $r_{j}$ and $z$ takes from $r_{j}$ to $r_{n+1}$ and we know that $xy^{i}z$ for $i\geq 0$ and $j\neq l$, so $y> 0$ and ${\color{Red} {l\leq p+1}}$, so $\left | xy \right |\leq p$
$a)$ Minimum pumping length $4.$ Because every string need to be divided in atleast $3$ parts $x,y,z.$ Otherwise it cannot be pumped.
$b)$ Minimum pumping length $1.$
$c)$ Minimum pumping length $4.$
$d)$ Minimum pumping length $3.$
$e)$ Minimum pumping length $2.$
$f)$Minimum pumping length $1.$
$g)$Minimum pumping length $3.$
$h)$Minimum pumping length $4.$
$i)$Minimum pumping length $5.$
$j)$Minimum pumping length $1.$
Ref:https://www.cs.auckland.ac.nz/~cristian/mfcsdir/cris/2010/tutorials/tut03_Solutions.pdf