$b^* = \{\varepsilon,b,bb,\ldots\}$ $a^*=\{\varepsilon,a,aa,\ldots\}$ we have to prove $(b^*a^*)^* = (b + a)^*$ and we know $b+a=a+b$
$b^*a^*=a^*$ if we take $b^*=\varepsilon$ string can start with $a$ u can see or $\varepsilon$
$b^*a^*=ba^*$ if we take $b^*=b$ string can start with $b$ u can see
$b^*a^*=ba^*$ if we take $b^*=bb$ string can start with $bb$ ucan see and so on $b^*a^*=bbbbbb.....a^*$ and if do the same thing with a then
$b^*a^*=b^*$ if $a^*=\varepsilon$ can ends with $b$ and $\varepsilon$ $b^*a^*=b^*a$ if $a^*=a$ can ends with $a$ and so on b$^*a^*=b^*.aaaaaaaaaa......$ $(b^*a^*)^*=(b^*a^*).(b^*a^*)$ now we have $(b^*a^*)^*$ we can say any string starts with $\varepsilon$ or $a$ or $b$ and ends with $\varepsilon$ or $a$ or $b$
and have any combination of $a$ or $b$ or both
so we can say that $(b^*a^*)^*=(b+a)^*$ or $(a+b)^*$
or go with this
now $b^*$ has $\varepsilon$ thats why when concatnated with $a^*$ ,$a$ can come first means if we take one value from $b^*$ which is $\varepsilon$ then $(\varepsilon.a^*)^*=(a^*)^* .......1$ and if we take $a^*=\varepsilon$ then
$(b^*\varepsilon)^*=(b^*)^* ........2$
and $(a^*)^*=a^*.a^*=a^*.(a+\varepsilon)^+$ ...from $1$ and $a^*=(a+\varepsilon)^+$