Dissolution of lead chloride because of the formation of- Na2PbO2 or [Pb(OH)4]-2?
Actually there are two reactions:


$$\ce{Pb(OH)2 + NaOH → Na[Pb(OH)3]}$$


  Lead(II) hydroxide react with sodium hydroxide to produce sodium
  trihydroxoplumbate(II). Sodium hydroxide - concentrated solution. (chemiday.com)

$$\ce{2NaOH + Pb(OH)2 → Na2[Pb(OH)4]}$$


  Sodium hydroxide react with lead(II) hydroxide to produce
  tetrahydroxoplumbate(II) sodium. (chemiday.com)



In the Wikipedia article of lead hydroxide, it is written that:


  In solution, lead(II) hydroxide is a somewhat weak base, forming
  lead(II) ion, $\ce{Pb^2+}$ under weakly acidic conditions. This cation
  hydrolyzes and under progressively increasing alkaline conditions,
  forms $\ce{Pb(OH)+}$, $\ce{Pb(OH)2(aq)}$, $\ce{Pb(OH)3−}$, and other
  polynuclear species like $\ce{Pb4(OH)4^4+}$, $\ce{Pb3(OH)4^2+}$,
  $\ce{Pb6O(OH)6^4+}$


Sodium plumbite ($\ce{Na2PbO2}$) is formed by the reaction of lead(II) oxide and sodium hydroxide.

$$\ce{PbO + 2NaOH -&gt; Na2PbO2 + H2O}$$



Some extra info.

The oxoions and hydroxoions of lead are called plumbates like  hydrated plumbate anion, $\ce{[Pb(OH)6]^2−}$ or the anhydrous anions $\ce{[PbO3]^2−}$
(meta-plumbate) or $\ce{[PbO4]^4-}$ (ortho-plumbate). The hydrated plumbates are formed by dissolving lead(IV) oxide in alkali and the anhydrous ones are formed by heating alkali or metal oxides with $\ce{PbO2}$. All plumbate(IV) salts are very strong oxidising agents and the hydrated plumbate(IV) salts decompose upon dehydration.